{"id":625585,"date":"2023-09-07T08:27:16","date_gmt":"2023-09-07T08:27:16","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=625585"},"modified":"2023-09-07T08:29:42","modified_gmt":"2023-09-07T08:29:42","slug":"kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/","title":{"rendered":"KC Sinha: Exercise 2.1 – Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901"},"content":{"rendered":"\n\n\n\n\n

(i) 3+\u221a5<\/strong>
Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

x=-\u221a5 , +\u221a5<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?
(i) x2<\/sup>=5<\/strong>
Sol :
\u21d2x=\u00b1\u221a5<\/p>\n\n\n\n

\u0905\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

(ii) $x^2=\\dfrac{16}{9}$<\/strong>
Sol :
\u21d2$x=\\sqrt{\\dfrac{16}{9}}$
\u21d2$x=\\pm \\dfrac{4}{3}$<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

(iii) $(x-1)^2=\\dfrac{49}{16}$<\/strong>
Sol :
\u21d2$x-1=\\sqrt{\\dfrac{49}{16}}$
\u21d2$x-1=\\pm \\dfrac{7}{4}$
\u21d2$x-1= +\\dfrac{7}{4}$ and $x-1=-\\dfrac{7}{4}$
\u21d2$x=\\dfrac{7}{4}+1$ and $x=-\\dfrac{7}{4}+1$
\u21d2$x=\\dfrac{7+4}{4}$ and $x=\\dfrac{-7+4}{4}$
\u21d2$x=\\dfrac{11}{4}$ and $x=\\dfrac{-3}{4}$<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

(iv) (x+1)(x-1)=0<\/strong>
Sol :
\u21d2x2<\/sup>-1=0
\u21d2x=\u00b1\u221a1<\/p>\n\n\n\n

\u21d2x=-1,+1<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

(v) $x^2=\\dfrac{19}{29}$<\/strong>
Sol :
\u21d2$x=\\sqrt{\\dfrac{19}{29}}$<\/p>\n\n\n\n

\u0905\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

(vi) (x-1)=5<\/strong>
Sol :
\u21d2x=5+1
\u21d2x=6<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

\u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0947 \u0932\u093f\u090f \u0926\u094b \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u093e \u0909\u0926\u093e\u0939\u0930\u0923 \u0926\u0947 \u091c\u093f\u0938\u0938\u0947 \u0909\u0928\u0915\u093e:
(i)<\/strong> \u092f\u094b\u0917\u092b\u0932 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964
Sol :
-\u221a3+\u221a3=0
-\u221a3,\u221a3<\/p>\n\n\n\n

(ii)<\/strong> \u092f\u094b\u0917\u092b\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964
Sol :<\/p>\n\n\n\n

2\u221a3+\u221a5=0<\/p>\n\n\n\n

2\u221a3,\u221a5<\/p>\n\n\n\n

(iii)<\/strong> \u0905\u0902\u0924\u0930 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

Sol :
\u221a3-\u221a3=0
\u221a3,\u221a3<\/p>\n\n\n\n

(iv)<\/strong> \u0905\u0902\u0924\u0930 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

Sol :
2\u221a3-\u221a3=\u221a3
2\u221a3,\u221a3<\/p>\n\n\n\n

(v)<\/strong> \u0917\u0941\u0923\u0928\u092b\u0932 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

Sol :
2\u221a3\u00d7\u221a3=6
2\u221a3,\u221a3<\/p>\n\n\n\n

(vi)<\/strong> \u0917\u0941\u0923\u0928\u092b\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964
Sol :
2\u221a3\u00d72\u221a2=4\u221a6
2\u221a3,2\u221a2<\/p>\n\n\n\n

(vii)<\/strong> \u092d\u093e\u0917\u092b\u0932 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964
Sol :
$\\frac{2\\sqrt{3}}{\\sqrt3}$
2\u221a3,\u221a3<\/p>\n\n\n\n

(viii)<\/strong> \u092d\u093e\u0917\u092b\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964
Sol :
$\\frac{4\\sqrt{6}}{\\sqrt{5}}$
4\u221a6,\u221a5<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

\u090f\u0915 \u092a\u0930\u093f\u092e\u0947\u092f \u090f\u0935\u0902 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0915\u093e \u0909\u0926\u093e\u0939\u0930\u0923 \u0926\u0947 \u091c\u093f\u0928\u0915\u093e \u0917\u0941\u0923\u0928\u092b\u0932 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u094b\u0924\u093e \u0939\u0948 \u0964
Sol :
\u21d20 , 2\u221a3
\u092a\u0930\u093f\u092e\u0947\u092f , \u0905\u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

\u21d20\u00d72\u221a3=0 \u092a\u0930\u093f\u092e\u0947\u092f<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0935\u094d\u092f\u0902\u091c\u0915\u094b \u092e\u0947 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947 :
(i)<\/strong> (5+\u221a5)(5-\u221a5)
Sol :
\u21d2(5)2<\/sup>-(\u221a5)2<\/sup>
\u21d225-5
\u21d220<\/p>\n\n\n\n

(ii)<\/strong> (5+\u221a7)(2+\u221a5)
Sol :
\u21d25(2+\u221a5)+\u221a7(2+\u221a5)
\u21d210+5\u221a5+2\u221a7+\u221a35<\/p>\n\n\n\n

(iii) (\u221a11-\u221a7)(\u221a11+\u221a7)<\/strong>
Sol :
Using identity:
(a+b)(a-b)=a2<\/sup>-b2<\/sup>
\u21d2(\u221a11)2<\/sup>-(\u221a7)2<\/sup>
\u21d211-7
\u21d24<\/p>\n\n\n\n

(iv) (11+\u221a11)(11-\u221a11)<\/strong>
Sol :
\u21d211(11-\u221a11)+\u221a11(11-\u221a11)
\u21d2121-11\u221a11+11\u221a11-11
\u21d2110<\/p>\n\n\n\n

(v) (3+\u221a2)(3-\u221a2)<\/strong>
Sol :
\u21d23(3-\u221a2)+\u221a2(3-\u221a2)
\u21d29-3\u221a2+3\u221a2-2
\u21d27<\/p>\n\n\n\n

(vi) (\u221a3+\u221a7)2<\/sup><\/strong>
Sol :
\u21d2(\u221a3)2<\/sup>+(\u221a7)2<\/sup>+2(\u221a3)(\u221a7)
\u21d23+7+2\u221a21
\u21d210+2\u221a21<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

(i) 5\u221a2+4\u221a2<\/strong>
Sol :
\u21d2\u221a2(5+4)
\u21d29\u221a2

(ii) 3\u221a7+2\u221a7<\/strong>
Sol :
\u21d2\u221a7(3+2)
\u21d25\u221a7
(iii) 8\u221a3-5\u221a3<\/strong>
Sol :
\u21d2\u221a3(8-5)
\u21d23\u221a3
(iv) 4\u221a7+5\u221a7-3\u221a7<\/strong>
Sol :
\u21d2\u221a7(4+5-3)
\u21d26\u221a7
(v) $8\\sqrt[3]{5}+7\\sqrt[3]{5}-13\\sqrt[3]{5}$<\/strong>
Sol :
\u21d2$\\sqrt[3]{5}(8+7-13)$
\u21d2$2\\sqrt[3]{5}$
(vi) 5\u221a3+2\u221a27<\/strong>
Sol :
\u21d25\u221a3+2\u221a3\u00d73\u00d73
\u21d2 5\u221a3+2\u00d73\u221a3
\u21d2 5\u221a3+6\u221a3
\u21d2\u221a3(6+5)
\u21d211\u221a3<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

(i) 4\u221a3-3\u221a2+2\u221a75<\/strong>
Sol :
\u21d24\u221a3-3\u221a2+2\u221a5\u00d75\u00d73
\u21d24\u221a3-3\u221a2+10\u221a3
\u21d24\u221a3+10\u221a3-3\u221a2
\u21d2\u221a3(4+10)-3\u221a2
\u21d2$14\\sqrt{3}-3\\sqrt{2}$
(ii) \u221a8+\u221a32-\u221a2<\/strong>
Sol :
\u21d2 \u221a2\u00d72\u00d72+\u221a2\u00d72\u00d72\u00d72\u00d72-\u221a2
\u21d22\u221a2+4\u221a2-\u221a2
\u21d22\u221a2+3\u221a2
\u21d25\u221a2
(iii) $\\sqrt{192}-\\dfrac{1}{2}\\sqrt{48}-\\sqrt{75}$<\/strong>
Sol :
\u21d2$\\sqrt{8\\times 8\\times 3}-\\dfrac{1}{2}\\sqrt{4\\times 4\\times 3}-\\sqrt{5\\times 5\\times 3}$
\u21d2$8\\sqrt{3}-\\dfrac{1}{2}\\times 4\\sqrt{3}-5\\sqrt{3}$
\u21d2$\\sqrt{3} \\left(8-\\dfrac{4}{2}-5\\right)$
\u21d2\u221a3(8-2-5)
\u21d2\u221a3<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u0947 \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 :
(i)<\/strong> (2\u221a2+5\u221a3)+(\u221a2-3\u221a3)<\/p>\n\n\n\n

Sol :
\u2066\u21d22\u221a2+5\u221a3+\u221a2-3\u221a3
\u21d23\u221a2+2\u221a3<\/p>\n\n\n\n

(ii)<\/strong> 6\u221a5\u00d72\u221a5<\/p>\n\n\n\n

Sol :
\u21d212\u00d75
\u21d260<\/p>\n\n\n\n

(iii)<\/strong> 8\u221a15\u00f72\u221a3<\/p>\n\n\n\n

Sol :
\u21d2$\\frac{8\\sqrt{15}}{2\\sqrt3}$<\/p>\n\n\n\n

\u21d24\u221a5<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0930\u0932\u0924\u092e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u093e\u0930\u0940 \u0917\u0941\u0923\u0915 \u0932\u093f\u0916\u0947:
(i)<\/strong> 5\u221a2
Sol :
\u221a2<\/p>\n\n\n\n

(ii)<\/strong> 2\u221a2
Sol :
\u221a2<\/p>\n\n\n\n

(iii)<\/strong> \u221a7
Sol :
\u221a7<\/p>\n\n\n\n

(iv)<\/strong> \u221a15
Sol :
\u221a15<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u092f\u0926\u093f a,b,c \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u092f\u093e\u090f\u0901 \u0939\u094b \u0924\u094b (i) $\\sqrt[5]{a^2b^3c^4}$ \u0924\u0925\u093e (ii) $\\sqrt[9]{a^2b^4c^8}$ \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u093e\u0930\u0940 \u0917\u0941\u0923\u0915 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
Sol :
(i) $\\sqrt[5]{a^3b^2c}$<\/p>\n\n\n\n

(ii) $\\sqrt[9]{a^7b^5c}$<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0947 \u0939\u0930 \u0915\u094b \u092a\u0930\u093f\u092e\u0947\u092f \u092c\u0928\u093e\u0915\u0930 \u0932\u093f\u0916\u0947:
(i)<\/strong> $\\frac{1}{\\sqrt2}$
Sol :
\u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u0930\u0923 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\frac{1}{\\sqrt{2}}\\times \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{1}{\\sqrt{12}}$
Sol :
$\\frac{1}{\\sqrt{12}}=\\frac{1}{\\sqrt{2\\times 2 \\times 3}}$<\/p>\n\n\n\n

$=\\frac{1}{2\\sqrt{3}}\\times \\frac{\\sqrt{3}}{\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{2\\times 3}=\\frac{\\sqrt{3}}{6}$<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{2\\sqrt{7}}{\\sqrt{11}}$
Sol :
$\\frac{2\\sqrt{7}}{\\sqrt{11}}\\times \\frac{\\sqrt{11}}{\\sqrt{11}}$<\/p>\n\n\n\n

$=\\frac{2\\sqrt{77}}{11}$<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{2}{\\sqrt{17}}$
Sol :<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

\u0930\u093f\u0915\u094d\u0924 \u0938\u094d\u0925\u093e\u0928\u094b \u092e\u0947 \u0939\u0930 \u0915\u094b \u092a\u0930\u093f\u092e\u0947\u092f \u092c\u0928\u093e\u0915\u0930 \u0932\u093f\u0916\u094b:
(i)<\/strong> $\\frac{1}{\\sqrt{2}+1}=\\dots$
Sol :
\u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u0930\u0923 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$\\frac{1}{\\sqrt{2}+1}\\times \\frac{\\sqrt{2}-1}{\\sqrt{2}-1}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}-1}{(\\sqrt{2})^2-1^2}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}-1}{2-1}=\\frac{\\sqrt{2}-1}{1}$<\/p>\n\n\n\n

=$\\sqrt{2}-1$<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{1}{2-\\sqrt{3}}=\\dots$
Sol :<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{3}{\\sqrt{5}+\\sqrt{3}}=\\dots$
Sol :<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{7}{\\sqrt{5}-\\sqrt{3}}=\\dots$
Sol :<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u0930\u0923\u093f\u092f\u094b \u0915\u094b \u0938\u0930\u0932\u0924\u092e \u0930\u0941\u092a \u092e\u0947 \u0932\u093f\u0916\u0947:
(i)<\/strong> \u221a48
Sol :
=\u221a2\u00d72\u00d72\u00d72\u00d73
=2\u00d72\u221a3
=4\u221a3<\/p>\n\n\n\n

(ii)<\/strong> \u221a175
Sol :<\/p>\n\n\n\n

(iii)<\/strong> \u221b72
Sol :
=\u221b2\u00d72\u00d72\u00d73\u00d73
=2\u00d7\u221b9 \u092f\u093e
=\u221b2\u00d72\u00d72\u00d73\u00d73\u00d73\u00d7$\\frac{1}{3}$
$=6\\sqrt[3]{\\frac{1}{3}}$<\/p>\n\n\n\n

(iv)<\/strong> \u221a125
Sol :<\/p>\n\n\n\n

(v)<\/strong> \u221b54
Sol :<\/p>\n\n\n\n

(vi)<\/strong> \u221b144
Sol :<\/p>\n\n\n\n

(vii)<\/strong> $\\sqrt[5]{320}$
Sol :<\/p>\n\n\n\n

(viii)<\/strong> $\\sqrt{\\frac{125}{63}}$
Sol :<\/p>\n\n\n\n

(ix)<\/strong> $\\sqrt{\\frac{112}{45}}$
Sol :
$=\\sqrt{\\frac{2\\times 2 \\times2 \\times 2\\times 7 }{3\\times 3 \\times 5}}$<\/p>\n\n\n\n

$=\\frac{4}{3}\\sqrt{\\frac{7}{5}}$<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0947 \u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u0930\u0923 \u0915\u0930\u0947 :
(i)<\/strong> $\\frac{1}{2+\\sqrt{3}}$
Sol :<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{1}{7+3\\sqrt{2}}$
Sol :
$=\\frac{1}{7+3\\sqrt{2}}\\times \\frac{7-3\\sqrt{2}}{7-3\\sqrt{2}}$<\/p>\n\n\n\n

$=\\frac{7-3\\sqrt{2}}{(7)^2-(3\\sqrt{2})^2}$<\/p>\n\n\n\n

$=\\frac{7-3\\sqrt{2}}{49-18}$<\/p>\n\n\n\n

$=\\frac{7-3\\sqrt{2}}{31}$<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{5}{\\sqrt{3}-\\sqrt{5}}$<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{6}{3\\sqrt{2}-2\\sqrt{3}}$
Sol :
$=\\frac{6}{3\\sqrt{2}-2\\sqrt{3}}\\times \\frac{3\\sqrt{2}+2\\sqrt{3}}{3\\sqrt{2}+2\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{6(3\\sqrt{2}+2\\sqrt{3})}{(3\\sqrt{2})^2-(2\\sqrt{3})^2}$<\/p>\n\n\n\n

$=\\frac{6(3\\sqrt{2}+2\\sqrt{3})}{18-12}$<\/p>\n\n\n\n

$=\\frac{6(3\\sqrt{2}+2\\sqrt{3})}{6}$<\/p>\n\n\n\n

=3\u221a2+2\u221a3<\/p>\n\n\n\n

(v)<\/strong> $\\frac{4}{\\sqrt{5}+\\sqrt{3}}$
Sol :<\/p>\n\n\n\n

(vi)<\/strong> $\\frac{5+\\sqrt{6}}{5-\\sqrt{6}}$
Sol :
$=\\frac{5+\\sqrt{6}}{5-\\sqrt{6}}\\times \\frac{5+\\sqrt{6}}{5+\\sqrt{6}}$<\/p>\n\n\n\n

$=\\frac{(5+\\sqrt{6})(5+\\sqrt{6}}{(5)^2-(\\sqrt{6}^2}$<\/p>\n\n\n\n

$=\\frac{25+5\\sqrt{6}+5\\sqrt{6}+6}{25-6}$<\/p>\n\n\n\n

$=\\frac{31+10\\sqrt{6}}{19}$<\/p>\n\n\n\n

(vii)<\/strong> $\\frac{3+\\sqrt{2}}{3-\\sqrt{2}}$<\/p>\n\n\n\n

(viii)<\/strong> $\\frac{2\\sqrt{6}-\\sqrt{5}}{3\\sqrt{5}-2\\sqrt{6}}$
Sol :
$=\\frac{2\\sqrt{6}-\\sqrt{5}}{3\\sqrt{5}-2\\sqrt{6}}\\times \\frac{3\\sqrt{5}+2\\sqrt{6}}{3\\sqrt{5}+2\\sqrt{6}}$<\/p>\n\n\n\n

$=\\frac{6\\sqrt{30}+4\\times 6-3\\times 5-2\\sqrt{30}}{45-24}$<\/p>\n\n\n\n

$=\\frac{4\\sqrt{30}+19}{21}$<\/p>\n\n\n\n

(ix)<\/strong> $\\frac{1+\\sqrt{2}}{\\sqrt{5}+\\sqrt{3}}$
Sol :<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:
(i)<\/strong> $\\frac{3}{5-\\sqrt{3}}+\\frac{2}{5+\\sqrt{3}}$
Sol :
\u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f \u0915\u0930\u0923 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$=\\frac{3}{5-\\sqrt{3}}\\times \\frac{5+\\sqrt{3}}{5+\\sqrt{3}}+\\frac{2}{5+\\sqrt{3}}\\times \\frac{5-\\sqrt{3}}{5-\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{3(5+\\sqrt{3})}{(5)^2-(\\sqrt{3})^2}+\\frac{2(5-\\sqrt{3})}{(5)^2-(\\sqrt{3})^2}$<\/p>\n\n\n\n

$=\\frac{3(5+\\sqrt{3})}{25-3}+\\frac{2(5-\\sqrt{3})}{25-3}$<\/p>\n\n\n\n

$=\\frac{3(5+\\sqrt{3})}{22}+\\frac{2(5-\\sqrt{3})}{22}$<\/p>\n\n\n\n

$=\\frac{3(5+\\sqrt{3}+2(5-\\sqrt{3}))}{22}$<\/p>\n\n\n\n

$=\\frac{15+3\\sqrt{3}+10-2\\sqrt{3}}{22}$<\/p>\n\n\n\n

$=\\frac{25+\\sqrt{3}}{22}$<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{\\sqrt{5}-2}{\\sqrt{5}+2}-\\frac{\\sqrt{5}+2}{\\sqrt{5}-2}$
Sol :
$=\\frac{(\\sqrt{5}-2)(\\sqrt{5}-2)-(\\sqrt{5}+2)(\\sqrt{5}+2)}{(\\sqrt{5}+2)(\\sqrt{5}-2)}$<\/p>\n\n\n\n

$=\\frac{(5-2\\sqrt{5}-2\\sqrt{5}+4)-(5+2\\sqrt{5}+2\\sqrt{5}+4)}{(\\sqrt{5})^2-(2)^2}$<\/p>\n\n\n\n

$=\\frac{(9-4\\sqrt{5})-(9+4\\sqrt{5})}{5-4}$<\/p>\n\n\n\n

$=\\frac{9-4\\sqrt{5}-9-4\\sqrt{5}}{1}$<\/p>\n\n\n\n

=-8\u221a5<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}+\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$
Sol :
$=\\frac{(\\sqrt{5}+\\sqrt{3})(\\sqrt{5}+\\sqrt{3})+(\\sqrt{5}-\\sqrt{3})(\\sqrt{5}-\\sqrt{3})}{(\\sqrt{5}-\\sqrt{3})(\\sqrt{5}+\\sqrt{3})}$<\/p>\n\n\n\n

$=\\frac{5+\\sqrt{15}+\\sqrt{15}+3+5-\\sqrt{15}-\\sqrt{15}+3}{(\\sqrt{5})^2-(\\sqrt{3})^2}$<\/p>\n\n\n\n

$=\\frac{16}{5-3}=\\frac{16}{2}=8$<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{1+\\sqrt{2}}{\\sqrt{5}+\\sqrt{3}}+\\frac{1-\\sqrt{2}}{\\sqrt{5}-\\sqrt{3}}$
Sol :
$=\\frac{(1+\\sqrt{2})(\\sqrt{5}-\\sqrt{3})+(1-\\sqrt{2})(\\sqrt{5}+\\sqrt{3})}{(\\sqrt{5}+\\sqrt{3})(\\sqrt{5}-\\sqrt{3})}$<\/p>\n\n\n\n

$=\\frac{(\\sqrt{5}-\\sqrt{3}+\\sqrt{10}-\\sqrt{6}+\\sqrt{5}+\\sqrt{3}-\\sqrt{10}-\\sqrt{6})()+()()}{(\\sqrt{5})^2-(\\sqrt{3})^2}$<\/p>\n\n\n\n

$=\\frac{2\\sqrt{5}-2\\sqrt{6}}{5-3}$<\/p>\n\n\n\n

$=\\frac{2(\\sqrt{6}-\\sqrt{6})}{2}$<\/p>\n\n\n\n

=\u221a5-\u221a6<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:<\/p>\n\n\n\n

(i)<\/strong> $\\frac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}+\\sqrt{5}}+\\sqrt{35}$
Sol :
\u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u0930\u0923 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$=\\frac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}+\\sqrt{5}}\\times \\frac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}-\\sqrt{5}}+\\sqrt{35}$<\/p>\n\n\n\n

$=\\frac{7-\\sqrt{35}-\\sqrt{35}+5}{(\\sqrt{7})^2-(\\sqrt{5})^2}+\\sqrt{35}$<\/p>\n\n\n\n

$=\\frac{12-2\\sqrt{35}}{7-5}+\\sqrt{35}$<\/p>\n\n\n\n

$=\\frac{2(6-\\sqrt{35})}{2}+\\sqrt{35}$<\/p>\n\n\n\n

=6-\u221a35+\u221a35<\/p>\n\n\n\n

=6<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}+2\\sqrt{6}$
Sol :
\u0939\u0930 \u0915\u093e \u092a\u0930\u093f\u092e\u0947\u092f\u0915\u0930\u0923 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}\\times \\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}+2\\sqrt{6}$<\/p>\n\n\n\n

$=\\frac{3-\\sqrt{6}-\\sqrt{6}+2}{(\\sqrt{3})^2-(\\sqrt{2})^2}+2\\sqrt{6}$<\/p>\n\n\n\n

$=\\frac{5-2\\sqrt{3}}{3-2}+2\\sqrt{6}$<\/p>\n\n\n\n

=5-2\u221a6+2\u221a6<\/p>\n\n\n\n

=5<\/p>\n\n\n\n

Question 18<\/h4>\n\n\n\n

(i)<\/strong> \u092f\u0926\u093f a=3+\u221a8 \u0924\u094b $a^2+\\frac{1}{a^2}$ \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
 Sol :
a=3+\u221a8<\/p>\n\n\n\n

$\\frac{1}{a}=\\frac{1}{3+\\sqrt{8}}$<\/p>\n\n\n\n

$=\\frac{1}{3+\\sqrt{8}}\\times \\frac{3-\\sqrt{8}}{3-\\sqrt{8}}$<\/p>\n\n\n\n

$=\\frac{3-\\sqrt{8}}{3^2-(\\sqrt{8})^2}$<\/p>\n\n\n\n

$=\\frac{3-\\sqrt{8}}{9-8}$<\/p>\n\n\n\n

$\\frac{1}{a}=3-\\sqrt{3}$<\/p>\n\n\n\n

$a^2+\\frac{1}{a^2}=\\left(a+\\frac{1}{a}\\right)^2-2.a.\\frac{1}{a}$<\/p>\n\n\n\n

$=(3+\\sqrt{8}+3-\\sqrt{8})^2-2$<\/p>\n\n\n\n

=(6)2<\/sup>-2<\/p>\n\n\n\n

=36-2<\/p>\n\n\n\n

=34<\/p>\n\n\n\n

(ii)<\/strong> \u092f\u0926\u093f $a=\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}$, $b=\\frac{\\sqrt{2}-1}{\\sqrt{2}+1}$ , \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f a2<\/sup>+b2<\/sup>+ab=35
Sol :
$a+b=\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}+\\frac{\\sqrt{2}-1}{\\sqrt{2}+1}$<\/p>\n\n\n\n

$=\\frac{(\\sqrt{2}+1)(\\sqrt{2}+1)+(\\sqrt{2}-1)(\\sqrt{2}-1)}{(\\sqrt{2}-1)(\\sqrt{2}+1)}$<\/p>\n\n\n\n

$=\\frac{2+\\sqrt{2}+\\sqrt{2}+1+2-\\sqrt{2}-\\sqrt{2}+1}{(\\sqrt{2})^2-(1)^2}$<\/p>\n\n\n\n

$=\\frac{6}{2-1}=6$<\/p>\n\n\n\n

L.H.S
a2<\/sup>+b2<\/sup>+ab=a2<\/sup>+b2<\/sup>+ab
=(a+b)2<\/sup>-2ab+ab
=(a+b)2<\/sup>-ab
$=6^2-\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}\\times \\frac{\\sqrt{2}-1}{\\sqrt{2}+1}$<\/p>\n\n\n\n

=36-1<\/p>\n\n\n\n

=35<\/p>\n\n\n\n

(iii)<\/strong> \u092f\u0926\u093f a=2+\u221a3 \u0924\u094b $a^3+\\frac{1}{a^3}$ \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964
Sol :
a=2+\u221a3<\/p>\n\n\n\n

$\\frac{1}{a}=\\frac{1}{2+\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{1}{2+\\sqrt{3}}\\times \\frac{2-\\sqrt{3}}{2-\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{2-\\sqrt{3}}{2^2-(\\sqrt{3})^2}$<\/p>\n\n\n\n

$\\frac{1}{a}=2-\\sqrt{3}$<\/p>\n\n\n\n

$a^3+\\frac{1}{a^3}=\\left(a+\\frac{1}{a}\\right)\\left(a^2-a.\\frac{1}{a}+\\frac{1}{a^2}\\right)$<\/p>\n\n\n\n

$=\\left(a+\\frac{1}{a}\\right)\\left((a+\\frac{1}{a})^2\\right)-2a.\\frac{1}{a}-1$<\/p>\n\n\n\n

$=4\\left[(4)^2-3\\right]$<\/p>\n\n\n\n

=4[16-3]<\/p>\n\n\n\n

=4\u00d713<\/p>\n\n\n\n

=52<\/p>\n\n\n\n

(iv)<\/strong> \u092f\u0926\u093f $x=\\frac{\\sqrt{3}+1}{2}$ \u0924\u094b 4x3<\/sup>+2x2<\/sup>-8x+7 \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
Sol :
$x^2=\\left(\\frac{\\sqrt{3}+1}{2}\\right)^2$<\/p>\n\n\n\n

$=\\frac{(\\sqrt{3}^2+2.\\sqrt{3}.1+1^2)}{4}$<\/p>\n\n\n\n

$=\\frac{4+2\\sqrt{3}}{4}=\\frac{2(2+\\sqrt{3})}{4}$<\/p>\n\n\n\n

$x^2=\\frac{2+\\sqrt{3}}{2}$<\/p>\n\n\n\n

$x^3=\\left(\\frac{\\sqrt{3}+1}{2}\\right)^2$<\/p>\n\n\n\n

$=\\frac{(\\sqrt{3})^3+3(\\sqrt{3})^2.1+3\\sqrt{3}.1^2+1^3}{8}$<\/p>\n\n\n\n

$=\\frac{3\\sqrt{3}+9+3\\sqrt{3}+1}{8}=\\frac{6\\sqrt{3}+10}{8}$<\/p>\n\n\n\n

$=\\frac{2(3\\sqrt{3}+5)}{8}$<\/p>\n\n\n\n

$=\\frac{3\\sqrt{3}+5}{4}$<\/p>\n\n\n\n

4x3<\/sup>+2x2<\/sup>-8x+7<\/p>\n\n\n\n

$=4\\left(\\frac{3\\sqrt{3}+5}{4}\\right)+2\\left(\\frac{2+\\sqrt{3}}{2}\\right)-8\\left(\\frac{\\sqrt{3}+1}{2}\\right)+7$<\/p>\n\n\n\n

=3\u221a3+5+2+\u221a3-4\u221a3-4+7<\/p>\n\n\n\n

=4\u221a3+10-4\u221a3<\/p>\n\n\n\n

=10<\/p>\n\n\n\n

(v)<\/strong> \u092f\u0926\u093f $a=\\frac{\\sqrt{3}+1}{\\sqrt{3}-1}$ \u0914\u0930 $b=\\frac{\\sqrt{3}-1}{\\sqrt{3}-1}$ \u0924\u094b a2<\/sup>-b2<\/sup>+ab \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
Sol :
a2<\/sup>+ab-b2<\/sup>=a2<\/sup>-b2<\/sup>+ab<\/p>\n\n\n\n

=(a+b)(a-b)+ab
……….<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

\u092f\u0926\u093f a \u0914\u0930 b \u0926\u094b \u092a\u0930\u093f\u0947\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 \u0939\u0948 \u0924\u094b \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0924\u093e\u0913 \u092e\u0947 a \u0914\u0930 b \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 :
(i)<\/strong> $\\frac{3+\\sqrt{7}}{3-\\sqrt{7}}=a+b\\sqrt{7}$
Sol :
$=\\frac{3+\\sqrt{7}}{3-\\sqrt{7}}\\times \\frac{3+\\sqrt{7}}{3+\\sqrt{7}}=a+b\\sqrt{7}$<\/p>\n\n\n\n

$\\frac{9+3\\sqrt{7}+3\\sqrt{7}+7}{(3)^2-(\\sqrt{7})^2}=a+b\\sqrt{7}$<\/p>\n\n\n\n

$\\frac{16+6\\sqrt{7}}{9-7}=a+b\\sqrt{7}$<\/p>\n\n\n\n

$\\frac{2(8+3\\sqrt{7})}{2}=a+b\\sqrt{7}$<\/p>\n\n\n\n

8+3\u221a7=a+b\u221a7<\/p>\n\n\n\n

\u092a\u0930\u093f\u092e\u0947\u092f \u0924\u0925\u093e \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0905\u0932\u0917-\u0905\u0932\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

a=8 ,<\/p>\n\n\n\n

b\u221a7=3\u221a7
$b=\\frac{3\\sqrt{7}}{\\sqrt{7}}$<\/p>\n\n\n\n

b=3<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{4+2\\sqrt{5}}{4-3\\sqrt{5}}=a+b\\sqrt{5}$
Sol :
$=\\frac{4+2\\sqrt{5}}{4-3\\sqrt{5}}\\times \\frac{4+3\\sqrt{5}}{4+3\\sqrt{5}}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\frac{16+12\\sqrt{5}+8\\sqrt{5}+30}{(4)^2-(3\\sqrt{5})^2}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\frac{46+20\\sqrt{5}}{16-45}=a+b\\sqrt{5}$<\/p>\n\n\n\n

${46+20\\sqrt{5}}{-29}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$-\\frac{46}{29}-\\frac{20}{29}\\sqrt{5}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\begin{array}{l|l}a=-\\frac{46}{29}& b\\sqrt{5}=\\frac{-20}{29}\\sqrt{5}\\\\&b=-\\frac{20}{29}\\end{array}$<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

\u092f\u0926\u093f $\\frac{5+\\sqrt{3}}{7-4\\sqrt{3}}=47a+\\sqrt{3}b$ \u0924\u094b a \u0914\u0930 b \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
Sol :
$\\frac{5+\\sqrt{3}}{7-4\\sqrt{3}}\\times \\frac{7+4\\sqrt{3}}{7+4\\sqrt{3}}=47a+\\sqrt{3}b$<\/p>\n\n\n\n

$\\frac{35+20\\sqrt{3}+7\\sqrt{3}+12}{(7)^2-(4\\sqrt{3})^2}=47a+\\sqrt{3}$<\/p>\n\n\n\n

$\\frac{47+27\\sqrt{3}}{49-48}=47a+\\sqrt{3}b$<\/p>\n\n\n\n

47+27\\sqrt{3}=47a+\\sqrt{3}b<\/p>\n\n\n\n

$\\begin{array}{l|l}47a=47&\\sqrt{3}b=27\\sqrt{3}\\\\a=\\frac{47}{47}=1&b=\\frac{27\\sqrt{3}}{\\sqrt{3}}\\\\&b=27\\end{array}$<\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0924\u093e\u0913 \u092e\u0947 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 a \u0914\u0930 b \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

(i)<\/strong> $\\frac{\\sqrt{5}-1}{\\sqrt{5}+1}+\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}=a+b\\sqrt{5}$
Sol :
$=\\frac{\\sqrt{5}-1}{\\sqrt{5}+1}+\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$=\\frac{(\\sqrt{5}-1)(\\sqrt{5}-1)+(\\sqrt{5}+1)(\\sqrt{5}+1)}{(\\sqrt{5}+1)(\\sqrt{5}-1)}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$=\\frac{5-\\sqrt{5}-\\sqrt{5}+1+5+\\sqrt{5}+\\sqrt{5}+1}{(\\sqrt{5})^2-1^2}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\frac{12}{5-1}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\frac{12}{4}=a+b\\sqrt{5}$<\/p>\n\n\n\n

$\\begin{array}{l|l}a=3&b\\sqrt{5}=0\\\\&b=\\frac{0}{\\sqrt{5}}=0\\end{array}$<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{7+\\sqrt{5}}{7-\\sqrt{5}}-\\frac{7+\\sqrt{5}}{7-\\sqrt{5}}=a+7\\sqrt{5}b$
Sol :<\/p>\n\n\n\n

Question 22<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u092e\u093e\u0928 \u090f\u0915 \u0927\u0928 \u092a\u0942\u0930\u094d\u0923\u093e\u0915 \u0915\u0947 \u0918\u093e\u0924 \u0915\u0947 \u0930\u0941\u092a \u092e\u0947 \u0932\u093f\u0916\u0947:
(i)<\/strong> 73<\/sup>.93<\/sup>
Sol : 633<\/sup><\/p>\n\n\n\n

(ii)<\/strong> 7-3<\/sup>.(9)-3<\/sup>
Sol : 63-3<\/sup>
$=\\frac{1}{(63)^3}$<\/p>\n\n\n\n

(iii)<\/strong> 172<\/sup>.175<\/sup>
Sol :
=172+5<\/sup>
=177<\/sup><\/p>\n\n\n\n

(iv)<\/strong> 172<\/sup>.17-5<\/sup>
=172<\/sup>-5<\/sup>
=17-3<\/sup><\/p>\n\n\n\n

(v)<\/strong> (52<\/sup>)7<\/sup>
=514<\/sup><\/p>\n\n\n\n

(vi)<\/strong> (52<\/sup>)-7<\/sup>
=5-14<\/sup><\/p>\n\n\n\n

(vii)<\/strong> $\\frac{23^{10}}{23^{7}}$
Sol :
=2310-7<\/sup>
=233<\/sup><\/p>\n\n\n\n

(viii)<\/strong> $\\frac{(23)^{-10}}{(23)^7}$
Sol :
=23-10-7<\/sup>
=23-17<\/sup>
$=\\frac{1}{(23)^{17}}$<\/p>\n\n\n\n

Question 23<\/h4>\n\n\n\n

\u0938\u0930\u0932 \u0915\u0940\u091c\u093f\u090f:
(i)<\/strong> $2^{\\frac{2}{3}}.2^{\\frac{1}{3}}$
Sol :
$=2^{\\frac{2}{3}+\\frac{1}{3}}$
$=2^{\\frac{2+1}{3}}$
$=2^{\\frac{3}{3}}=2$<\/p>\n\n\n\n

(ii)<\/strong> $\\left(3^{\\frac{1}{5}}\\right)^4$
Sol :
$=3^{\\frac{1}{5}\\times 4}$
$=3^{\\frac{4}{5}}$<\/p>\n\n\n\n

(iii)<\/strong> $13^{\\frac{1}{5}.17^{\\frac{1}{5}}}$
Sol :
$=(13\\times 17)^{\\frac{1}{5}}$
$=(221)^{\\frac{1}{5}}$<\/p>\n\n\n\n

(iv)<\/strong> $\\dfrac{7^{\\frac{1}{5}}}{7^{\\frac{1}{3}}}$
Sol :
$=7^{\\frac{1}{5}-\\frac{1}{3}}$
$=7^{\\frac{3-5}{15}}$
$=7^{\\frac{-2}{15}}=\\frac{1}{7^{\\frac{2}{15}}}$<\/p>\n\n\n\n

Question 24<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:
(i)<\/strong> \u221a15\u00d7\u221a7
Sol :
\u221a15\u00d77
=\u221a105<\/p>\n\n\n\n

(ii)<\/strong> \u221b18\u00d7\u221b15
Sol :
=\u221b18\u00d715
=\u221b2\u00d73\u00d73\u00d73\u00d75
=3\u221b10<\/p>\n\n\n\n

(iii)<\/strong> \u221c5\u00d7\u221c8
Sol :
=\u221c5\u00d78
=\u221c40
=\u221c5\u00d72\u00d72\u00d72\u00d72\u00d7$\\frac{1}{2}$
$=2\\sqrt[4]{\\frac{5}{2}}$<\/p>\n\n\n\n

(iv)<\/strong> $\\sqrt[7]{9}\\times \\sqrt[7]{5}\\times \\sqrt[7]{2}$
Sol :
$=\\sqrt[7]{9\\times 5 \\times 2}$
$=\\sqrt[7]{90}$<\/p>\n\n\n\n

(v)<\/strong> $\\sqrt[8]{12} \\div \\sqrt[8]{3}$
Sol :
$=\\frac{\\sqrt[8]{12}}{\\sqrt[8]{3}}$
$=\\sqrt[8]{\\frac{12}{3}}$
$=\\sqrt[8]{4}$<\/p>\n\n\n\n

(vi)<\/strong> $\\sqrt[5]{24} \\div \\sqrt[5]{6}$
Sol :
$=\\frac{\\sqrt[5]{24}}{\\sqrt[5]{6}}$
$=\\sqrt[5]{\\frac{24}{6}}$
$=\\sqrt[5]{4}$<\/p>\n\n\n\n

Question 25<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0938\u0930\u0932 \u0915\u0930\u0947:
(i)<\/strong> \u221b2\u00d7\u221a5
Sol :
$=2^{\\frac{1}{3}}\\times 5^{\\frac{1}{2}}$
$=2^{\\frac{1}{3}\\times \\frac{1}{2}\\times 2}\\times 5^{\\frac{1}{2}\\times \\frac{1}{3}\\times 3}$
$=(2^2)^{\\frac{1}{6}}\\times (5^{3})^{\\frac{1}{6}}$
$=(4)^{\\frac{1}{6}}\\times(125)^{\\frac{1}{6}}$
$=(500)^{\\frac{1}{6}}$
$=\\sqrt[6]{500}$<\/p>\n\n\n\n

(ii)<\/strong> \u221b7\u00d7\u221a2
Sol :<\/p>\n\n\n\n

(iii)<\/strong> \u221b5\u00d7\u221a3
Sol :<\/p>\n\n\n\n

(iv)<\/strong> \u221b7\u00d7\u221c3
Sol :
$=7^{\\frac{1}{3}} \\times 3^{\\frac{1}{4}}$
$=7^{\\frac{1}{3}\\times \\frac{1}{4}\\times 4} \\times 3^{\\frac{1}{4}\\times \\frac{1}{3}\\times 3}$
$=(7^{4})^{\\frac{1}{12}}\\times (3^3)^{\\frac{1}{12}}$
$=(64827)^{\\frac{1}{12}}$
$=\\sqrt[12]{64827}$<\/p>\n\n\n\n

(v)<\/strong> \u221a2.\u221b3.\u221c4
Sol :
$=2^{\\frac{1}{2}}\\times 3^{\\frac{1}{3}}\\times 4^{\\frac{1}{4}}$
$=2^{\\frac{1}{2}}\\times 3^{\\frac{1}{3}}\\times 4^{\\frac{1}{4}}$
$=2^{\\frac{1}{2}\\times \\frac{1}{6}\\times 6}\\times 3^{\\frac{1}{3}\\times \\frac{1}{4}\\times 4}\\times 4^{\\frac{1}{4}\\times \\frac{1}{3}\\times 3}$
$=(2^6)^{\\frac{1}{12}}\\times (3^4)^{\\frac{1}{12}}\\times (4^{3})^{\\frac{1}{12}}$
$=(64)^{\\frac{1}{12}}\\times (81)^{\\frac{1}{12}}\\times (64)^{\\frac{1}{12}}$
$=(331776)^{\\frac{1}{12}}$
$=\\sqrt[12]{331776}$<\/p>\n\n\n\n

(vi)<\/strong> \u221a3.\u221b4.\u221c5
Sol :<\/p>\n\n\n\n

(vii)<\/strong> 24\u00f7 \u221b200
Sol :
$\\frac{24}{\\sqrt[3]{200}}$
$=\\frac{24}{\\sqrt[3]{\\frac{2\\times 2\\times 2\\times 5\\times 5}{12}}}$
$=\\frac{24}{2\\times \\sqrt[3]{25}}$
$=\\frac{12}{\\sqrt[3]{25}}$
$=\\frac{12}{25^{\\frac{1}{3}}}$
$=\\frac{12^{frac{1}{3}\\times 3}}{25^{\\frac{1}{3}}}$
$=\\frac{(12^3)^{\\frac{1}{3}}}{25^{\\frac{1}{3}}}$
$=\\left(\\frac{1728}{25}\\right)^{\\frac{1}{3}}$
$=\\sqrt[3]{\\frac{1728}{25}}$<\/p>\n\n\n\n

(viii)<\/strong> \u221c36\u00f7\u221b6
Sol :
$=\\frac{\\sqrt[4]{36}}{\\sqrt[4]{6}}$
$=\\frac{(36)^{\\frac{1}{4}}}{6^{\\frac{1}{3}}}$
$=\\dfrac{6^{2\\times \\frac{1}{4}}}{6^{\\frac{1}{3}}}$
$=\\frac{6^{\\frac{1}{6}}}{6^{\\frac{1}{3}}}$
$=6^{\\frac{1}{2}-\\frac{1}{3}}$
$=6^{\\frac{3-2}{6}}$
$=6^{\\frac{1}{6}}$<\/p>\n\n\n\n

Question 26<\/h4>\n\n\n\n

\u092f\u0926\u093f \u221a2=1.414 , \u221a3=1.732 , \u221a5=2.236 \u0924\u0925\u093e \u221a10=3.162  \u0924\u094b \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u0947 \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947:
(i)<\/strong> $\\frac{1}{\\sqrt{2}}$
Sol :
$=\\frac{1}{\\sqrt{2}}\\times \\frac{\\sqrt{2}}{\\sqrt{2}}$
$=\\frac{\\sqrt{2}}{2}=\\frac{1.414}{2}$
=0.707<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{3}{\\sqrt{10}}$
Sol :<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{2+\\sqrt{3}}{3}$
Sol :<\/p>\n\n\n\n

(iv)<\/strong> $\\frac{\\sqrt{10}+\\sqrt{15}}{\\sqrt{2}}$
Sol :<\/p>\n\n\n\n

Question 27<\/h4>\n\n\n\n

\u092f\u0926\u093f m \u0914\u0931 n \u0926\u094b \u092a\u094d\u0930\u093e\u0915\u0943\u0924 \u0938\u0902\u0916\u094d\u092f\u093e\u090f\u0901 \u0939\u094b \u0924\u093e\u0915\u093f mn<\/sup>=25 \u0924\u094b nm<\/sup> \u0915\u093e \u092e\u093e\u0928 \u0939\u094b\u0917\u093e:
(i) 4
(ii) 10
(iii) 32
(iv) 16
Sol :
\u21d2mn<\/sup>=25
\u21d2mn<\/sup>=52<\/sup>
\u21d2nm<\/sup>=25<\/sup>=32<\/p>\n\n\n\n

Question 28<\/h4>\n\n\n\n

$\\sqrt{10}\\times \\sqrt{15}$ \u092c\u0930\u093e\u092c\u0930 \u0939\u0948
(i)<\/strong> 5\u221a6
(ii)<\/strong> 6\u221a5
(iii)<\/strong> \u221a30
(iv)<\/strong> \u221a25<\/p>\n\n\n\n

Sol :
=\u221a10\u00d7\u221a15
=\u221a2\u00d75\u00d73\u00d75
=5\u221a6<\/p>\n\n\n\n

Question 29<\/h4>\n\n\n\n

$\\sqrt[5]{6}\\times \\sqrt[5]{6^{0}}$ \u092c\u0930\u093e\u092c\u0930 \u0939\u0948
(i)<\/strong> $\\sqrt[5]{36}$
(ii)<\/strong> $\\sqrt[5]{6\\times 0}$
(iii)<\/strong> $\\sqrt[5]{6}$
(iv)<\/strong> $\\sqrt[5]{12}$<\/p>\n\n\n\n

Sol :
$=\\sqrt[5]{6}\\times \\sqrt[5]{6^0}$
$=\\sqrt[5]{6}$<\/p>\n\n\n\n

Question 30<\/h4>\n\n\n\n

$\\sqrt[3]{8^2}$ \u092c\u0930\u093e\u092c\u0930 \u0939\u0948
(i)<\/strong> $8^{\\frac{2}{3}}$
(ii)<\/strong> $8^{\\frac{3}{2}}$
(iii)<\/strong> $4\\times 4^{\\frac{2}{3}}$
(iv)<\/strong> 4
Sol :
$=\\sqrt[3]{8^2}=8^{2\\times \\frac{1}{3}}$
$=8^{\\frac{2}{3}}$
$=2^{3\\times \\frac{2}{3}}$
$=2^2=4$<\/p>\n\n\n\n

Question 31<\/h4>\n\n\n\n

\u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f \u221a2+\u221a3 \u090f\u0915 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948
Sol :
\u092e\u093e\u0928\u093e \u221a2+\u221a3 \u090f\u0915 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

$\\sqrt{2}+\\sqrt{3}=\\frac{a}{b}$ (\u091c\u0939\u093e\u0901 a \u0914\u0930 b \u092a\u0942\u0930\u094d\u0923\u093e\u0902\u0915 \u0939\u0948 \u0964)<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0924\u0930\u092b \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

$(\\sqrt{2}+\\sqrt{3})^2=\\frac{a^2}{b^2}$<\/p>\n\n\n\n

$(\\sqrt{2})^2+(\\sqrt{3})^2+2.\\sqrt{2}.\\sqrt{3}=\\frac{a^2}{b^2}$<\/p>\n\n\n\n

$2\\sqrt{6}=\\frac{a^2}{b^2}-5$<\/p>\n\n\n\n

$2\\sqrt{6}=\\frac{a^2-5b^2}{b^2}$<\/p>\n\n\n\n

$\\sqrt{6}=\\frac{a^2-5b^2}{2b^2}$<\/p>\n\n\n\n

R.H.S \u092e\u0947 a,b, 2 \u0924\u0925\u093e 5 \u090f\u0915 \u092a\u0942\u0930\u094d\u0923\u093e\u0902\u0915 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2234$\\frac{a^2-5b^2}{2b^2}$ \u090f\u0915 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

\u221a6 \u090f\u0915 \u092a\u0930\u093f\u092e\u0947\u092f \u0939\u094b\u0917\u093e \u0964<\/p>\n\n\n\n

\u0932\u0947\u0915\u093f\u0928 \u0939\u092e\u093e\u0930\u0940 \u0905\u0935\u0927\u093e\u0930\u0923\u093e \u0917\u0932\u0924 \u0915\u0940 \u221a6 \u090f\u0915 \u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

\u0905\u0924: \u221a2+\u221a3 \u090f\u0915 \u0905\u092a\u0930\u093f\u092e\u0947\u092f \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii) $x^2=\\dfrac{16}{9}$Sol :\u21d2$x=\\sqrt{\\dfrac{16}{9}}$\u21d2$x=\\pm \\dfrac{4}{3}$ \u092a\u0930\u093f\u092e\u0947\u092f (iii) $(x-1)^2=\\dfrac{49}{16}$Sol :\u21d2$x-1=\\sqrt{\\dfrac{49}{16}}$\u21d2$x-1=\\pm \\dfrac{7}{4}$\u21d2$x-1= +\\dfrac{7}{4}$ and $x-1=-\\dfrac{7}{4}$\u21d2$x=\\dfrac{7}{4}+1$ and $x=-\\dfrac{7}{4}+1$\u21d2$x=\\dfrac{7+4}{4}$ and $x=\\dfrac{-7+4}{4}$\u21d2$x=\\dfrac{11}{4}$ and $x=\\dfrac{-3}{4}$ \u092a\u0930\u093f\u092e\u0947\u092f (iv) (x+1)(x-1)=0Sol :\u21d2x2-1=0\u21d2x=\u00b1\u221a1 \u21d2x=-1,+1 \u092a\u0930\u093f\u092e\u0947\u092f (v) $x^2=\\dfrac{19}{29}$Sol :\u21d2$x=\\sqrt{\\dfrac{19}{29}}$ \u0905\u092a\u0930\u093f\u092e\u0947\u092f (vi) (x-1)=5Sol […]<\/p>\n","protected":false},"author":302,"featured_media":625587,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[921],"tags":[],"boards":[],"class_list":["post-625585","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901\" \/>\n<meta property=\"og:description\" content=\"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-07T08:27:16+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-07T08:29:42+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 2.1 – Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901\",\"datePublished\":\"2023-09-07T08:27:16+00:00\",\"dateModified\":\"2023-09-07T08:29:42+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\"},\"wordCount\":1774,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg\",\"articleSection\":[\"class 9\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\",\"name\":\"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901 - IndCareer Schools\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg\",\"datePublished\":\"2023-09-07T08:27:16+00:00\",\"dateModified\":\"2023-09-07T08:29:42+00:00\",\"description\":\"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)\",\"breadcrumb\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg\",\"width\":1600,\"height\":901,\"caption\":\"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.indcareer.com\/schools\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"class 9\",\"item\":\"https:\/\/www.indcareer.com\/schools\/class-9\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"KC Sinha: Exercise 2.1 – Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"name\":\"IndCareer Schools\",\"description\":\"School Admissions & Notices\",\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\",\"name\":\"IndCareer\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"width\":512,\"height\":250,\"caption\":\"IndCareer\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/indcareer\",\"https:\/\/x.com\/indcareer\",\"https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ\"],\"email\":\"info@ebharat.in\",\"legalName\":\"IndCareer\",\"numberOfEmployees\":{\"@type\":\"QuantitativeValue\",\"minValue\":\"1\",\"maxValue\":\"10\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\",\"name\":\"Pooja\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"caption\":\"Pooja\"}}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901 - IndCareer Schools","description":"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901","og_description":"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-07T08:27:16+00:00","article_modified_time":"2023-09-07T08:29:42+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 2.1 – Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901","datePublished":"2023-09-07T08:27:16+00:00","dateModified":"2023-09-07T08:29:42+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/"},"wordCount":1774,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg","articleSection":["class 9"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/","name":"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg","datePublished":"2023-09-07T08:27:16+00:00","dateModified":"2023-09-07T08:29:42+00:00","description":"(i) 3+\u221a5Sol : \u0905\u092a\u0930\u093f\u092e\u0947\u092f x=-\u221a5 , +\u221a5 \u092a\u0930\u093f\u092e\u0947\u092f Question 3 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923\u094b \u092e\u0947\u0902 \u0915\u093f\u0938\u0915\u093e \u0939\u0932 \u0905\u092a\u0930\u093f\u092e\u0947\u0902\u092f \u0938\u0902\u0916\u094d\u092f\u093e\u0913 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u0947 \u0939\u0948 ?(i) x2=5Sol :\u21d2x=\u00b1\u221a5 \u0905\u092a\u0930\u093f\u092e\u0947\u092f (ii)","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-1-2-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 2.1 - Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901"},{"@type":"BreadcrumbList","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-1-mathematics-solution-class-9-chapter-2-vastavik-sankhyao-par-sankriyaein\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.indcareer.com\/schools\/"},{"@type":"ListItem","position":2,"name":"class 9","item":"https:\/\/www.indcareer.com\/schools\/class-9\/"},{"@type":"ListItem","position":3,"name":"KC Sinha: Exercise 2.1 – Mathematics Solution Class 9 Chapter 2 \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u092a\u0930 \u0938\u0902\u0915\u094d\u0930\u093f\u092f\u093e\u090f\u0901"}]},{"@type":"WebSite","@id":"https:\/\/www.indcareer.com\/schools\/#website","url":"https:\/\/www.indcareer.com\/schools\/","name":"IndCareer Schools","description":"School Admissions & Notices","publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/www.indcareer.com\/schools\/#organization","name":"IndCareer","url":"https:\/\/www.indcareer.com\/schools\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","width":512,"height":250,"caption":"IndCareer"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/indcareer","https:\/\/x.com\/indcareer","https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ"],"email":"info@ebharat.in","legalName":"IndCareer","numberOfEmployees":{"@type":"QuantitativeValue","minValue":"1","maxValue":"10"}},{"@type":"Person","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e","name":"Pooja","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","caption":"Pooja"}}]}},"_links":{"self":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/625585","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/users\/302"}],"replies":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/comments?post=625585"}],"version-history":[{"count":0,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/625585\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media\/625587"}],"wp:attachment":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media?parent=625585"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/categories?post=625585"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/tags?post=625585"},{"taxonomy":"boards","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/boards?post=625585"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}