{"id":624304,"date":"2023-09-01T08:02:25","date_gmt":"2023-09-01T08:02:25","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624304"},"modified":"2023-09-04T12:38:42","modified_gmt":"2023-09-04T12:38:42","slug":"kc-sinha-exercise-7-3-mathematics-solution-class-10-chapter-7-quadratic-equations","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-7-3-mathematics-solution-class-10-chapter-7-quadratic-equations\/","title":{"rendered":"KC Sinha: Exercise 7.3 – Mathematics Solution Class 10 Chapter 7 Quadratic Equations"},"content":{"rendered":"\n\n\n\n\n

Question 1<\/h4>\n\n\n\n

Determine whether <\/strong>$x=\\frac{3}{2}$ <\/strong>and $x=-\\frac{4}{3}$ <\/strong>are the solutions of the equation 6x2<\/sup>-x-12=0 or not.<\/p>\n\n\n\n

Sol :
Put both the values of x in the equation.
When $x=\\frac{3}{2}$
$6\\left(\\frac{3}{2}\\right)^{2}-\\frac{3}{2}-12=0$
$6 \\times \\frac{9}{4}-\\frac{3}{2}-12=0$
$\\frac{54-48-6}{4}$
$\\frac{54-54}{4}$
= 0
<\/p>\n\n\n\n

When $x=-\\frac{4}{3}$
$6\\left(-\\frac{4}{3}\\right)^{2}-\\frac{4}{3}-12=0$
$6\\left(\\frac{16}{9}\\right)+\\frac{4}{3}-12=0$
$\\frac{96+12-108}{9}$
$\\frac{108-108}{9}$
= 0
R.H.S = L.H.S
Therefore, $\\mathrm{x}=\\frac{3}{2}$ and $x=-\\frac{4}{3}$ are the solutions of the given equation.<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

Determine whether (i) x = 1, (ii) x = 3 are the solutions of the equation x2<\/sup> \u2014 5x + 4 = 0 or not.<\/strong><\/p>\n\n\n\n

Sol :
Put both the values of x in the equation.
When x = 1
12<\/sup> \u2013 5(1) + 4
1 \u2013 5 + 4
= 0
Therefore, it is the solution to the equation.
When x = 3
32<\/sup> \u2013 5(3) + 4 = 0
9 \u2013 15 + 4
= \u20132
Therefore, it is not the solution to the equation.<\/p>\n\n\n\n

\n\n\n\n

Question 3 <\/h4>\n\n\n\n

Determine whether x = \u221a3 and x = \u20142\u221a3 are solutions of the equation x2<\/sup> \u2013 3\u221a3x + 6 = 0<\/strong><\/p>\n\n\n\n

Sol :
Put both the values of x in the equation.
When x = \u221a3
(\u221a3)2<\/sup> \u20133\u221a3(\u221a3) + 6 = 0
3 \u2013 (3)3 + 6
3 \u2013 9 + 6
9 \u2013 9
= 0
Therefore, it is the solution to the equation.
When x = \u20132\u221a3
(\u20132\u221a3)2<\/sup> \u20133\u221a3(\u20132\u221a3) + 6 = 0
12 + 18 + 6
= 36
Therefore, it is not the solution to the equation.<\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

For 2x2<\/sup>\u2014 5x \u20143 = 0, determine which of the following are solutions?<\/strong><\/p>\n\n\n\n

(i) x = 3 <\/strong><\/p>\n\n\n\n

(ii) x= \u20132<\/strong>
(iii)$x=-\\frac{1}{2}$ <\/strong><\/p>\n\n\n\n

(iv) <\/strong>$x=-\\frac{1}{3}$<\/strong>
Sol :
The two possible solutions are x = 3 and $=-\\frac{1}{2}$
<\/p>\n\n\n\n

Since this question is given in standard form, meaning that it follows the form: ax2<\/sup> + by + c = 0, we can use the quadratic formula to solve for x:
$x=\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$x=\\frac{-(-5) \\pm \\sqrt{(-5)^{2}-4(2)(-3)}}{2 \\times 2}$
$x=\\frac{5 \\pm \\sqrt{25+24}}{4}$
$x=\\frac{5 \\pm 7}{4}$
$x=\\frac{5+7}{4}, x=\\frac{5-7}{4}$
$\\mathrm{x}=\\frac{12}{4}, \\mathrm{x}=\\frac{-2}{4}$
x = 3 , $x=-\\frac{1}{2}$
That value of x is correct as well!
Therefore, the two possible solutions are:
x=3
x=\u22120.50<\/p>\n\n\n\n

\n\n\n\n

Question 5 <\/h4>\n\n\n\n

Determine whether (i) x= \u221a2, (ii) x = \u20132\u221a2 are the solutions of the equation x2<\/sup> + \u221a2 x \u2013 4 = 0 or not.<\/strong><\/p>\n\n\n\n

Sol :
Put x = \u221a2
(\u221a2)2<\/sup> + \u221a2(\u221a2) \u2013 4 = 0
2 + 2 \u2013 4
4 \u2013 4
= 0
Therefore, it is the solution to the equation.
When x = \u20132\u221a2
(\u20132\u221a2)2<\/sup> + \u221a2(\u20132\u221a2) \u2013 4 = 0
8 \u2013 4 \u2013 4
= 0
Therefore, it is the solution to the equation.<\/p>\n\n\n\n

\n\n\n\n

Question 6 <\/h4>\n\n\n\n

Show that x = \u2014 3 is a solution of x2<\/sup> + 6x + 9 = 0.<\/strong><\/p>\n\n\n\n

Sol :
Put x = \u20133 in the equation.
(\u20133)2<\/sup> + 6(\u20133) + 9
9 \u2013 18 + 9
=0
Hence it is a solution<\/p>\n\n\n\n

\n\n\n\n

Question 7 <\/h4>\n\n\n\n

Show that x = \u2014 3 is a solution of 2x2<\/sup> + 5x \u20133 = 0.<\/strong><\/p>\n\n\n\n

Sol :
Put x = \u20133 in the equation.
2(\u20133)2<\/sup> + 5(\u20133) \u20133 = 0
18 \u2013 15 \u20133
18 \u2013 18
= 0
Therefore, x = \u20133 is the solution of the equation.<\/p>\n\n\n\n

\n\n\n\n

Question 8 <\/h4>\n\n\n\n

Show that x = \u2014 2 is a solution of 3x2<\/sup> + 13x + 14 = 0.<\/strong><\/p>\n\n\n\n

Sol :
The given quadratic equation is 3x2<\/sup> + 13x + 14 = 0
Putting x = \u2013 2,
L.H.S.
3.(\u20132)2<\/sup> + 13.(\u20132) + 14
3 x 4 \u2013 26 + 14
12 \u2013 26 + 14
26 \u2013 26
= 0
Hence, x = \u2013 2 is a solution of 3x + 13x + 14 = 0<\/p>\n\n\n\n

\n\n\n\n

Question 9 <\/h4>\n\n\n\n

For what value of k, <\/strong>$x=\\frac{2}{3}$ is the solution of the equation<\/strong><\/p>\n\n\n\n

kx2<\/sup> \u2013 x \u2013 2 = 0.<\/strong>
Sol <\/p>\n\n\n\n

$x=\\frac{2}{3}$
kx2<\/sup> \u2013 x \u2013 2 = 0
$k\\left(\\frac{2}{3}\\right)^{2}-\\frac{2}{3}-2$
$\\frac{4 \\mathrm{k}}{9}-\\frac{2}{3}-2$
$\\frac{4 \\mathrm{k}-6-18}{9}$
4k \u2013 24 = 0
4k = 24
k = 6<\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

For what value of k, <\/strong>$\\mathrm{x}=-\\frac{1}{2}$ is a solution of the equation 3x<\/strong>2<\/sup> + 2kx \u2014 3 = 0<\/strong><\/p>\n\n\n\n

Sol :
Put the value of x in the equation.
3x2<\/sup> + 2kx \u2014 3 = 0
$3\\left(-\\frac{1}{2}\\right)^{2}+2 \\mathrm{k}\\left(-\\frac{1}{2}\\right)-3=0$
$\\frac{3}{4}-\\frac{2 \\mathrm{k}}{2}-3=0$
$\\frac{3-4 k-12}{4}=0$
\u20139 \u2013 4k = 0
\u20134k = 9
$\\mathrm{k}=-\\frac{9}{4}$<\/p>\n\n\n\n

\n\n\n\n

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

For what values of a and b, <\/strong>$x=\\frac{3}{4}$ and x = \u2014 2 are solutions of the equation ax<\/strong>2<\/sup> + bx \u2014 6 = 0.<\/strong><\/p>\n\n\n\n

Sol :
Put x = 3\/4
ax2<\/sup> + bx \u2014 6 = 0
$a\\left(\\frac{3}{4}\\right)^{2}+b\\left(\\frac{3}{4}\\right)-6=0$
$\\frac{9 a}{16}+\\frac{3 b}{4}-5=0$
$\\frac{9 a+12 b-96}{16}=0$
9a + 12b \u2013 96 = 0 divide by 3
3a + 4b \u2013 32 = 0
3a + 4b = 32 (1)
<\/p>\n\n\n\n

Put x = \u20132
ax2<\/sup> + bx \u2014 6 = 0
a(\u20132)2<\/sup> + b(\u20132) \u2013 6 = 0
4a \u2013 2b \u2013 6 = 0
4a \u2013 2b = 6 (2)
<\/p>\n\n\n\n

Eliminate (1) and (2)
3a + 4b = 32
4a \u2013 2b = 6 \u00d72<\/p>\n\n\n\n

$\\begin{aligned}3a+4b&=32\\\\8a-4b&=12\\\\ \\hline 11a&=44\\end{aligned}$<\/p>\n\n\n\n

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

Put a = 4 in equation (1).
3a + 4b = 32
3(4) + 4b = 32
12 + 4b = 32
4b = 32 \u2013 12
4b = 20
b = 5<\/p>\n\n\n\n

\n\n\n\n

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

For what value of k, x = a is a solution of the equation<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 (a + b) x + k = 0.<\/strong>
Sol :
x2<\/sup> \u2013 (a + b)x + k = 0
Put x = a
a2<\/sup> \u2013 (a + b) a + k
a2<\/sup> \u2013 a2<\/sup> + ab + k = 0
ab + k = 0
k = \u2013ab<\/p>\n\n\n\n

\n\n\n\n

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

Determine the value of k, a and b in each of the following quadratic equation, for which the given value of x is the root of the given quadratic equation:<\/strong><\/p>\n\n\n\n

(i) kx2<\/sup> \u2014 5x + 6 = 0 ; x = 2<\/strong>
(ii) 6x2<\/sup> + kx \u2014 \u221a6 = 0; <\/strong>$x=-\\frac{\\sqrt{3}}{2}$<\/strong>
(iii) ax2<\/sup> \u2014 13x + b = 0; x =2 and x = \u20142 find a, b<\/strong>
(iv) ax2<\/sup>+ bx \u2013 10 = 0; <\/strong>$x=-\\frac{2}{5}$<\/strong> and <\/strong>$x=\\frac{5}{3}$<\/strong><\/p>\n\n\n\n

Sol :
(i) kx2<\/sup> \u2013 5x + 6 = 0
Put x = 2
(ii) k2<\/sup> \u2013 5(2) + 6 = 0
4k \u2013 10 + 6 = 0
4k = 4
k = 1
<\/p>\n\n\n\n

(iii) 6x2<\/sup> + kx -\u221a6 = 0
Put x = \u2013 2\/\u221a3
$6\\left(-\\frac{\\sqrt{3}}{2}\\right)^{2}+\\mathrm{k}\\left(-\\frac{\\sqrt{3}}{2}\\right)-\\sqrt{6}=0$<\/p>\n\n\n\n

$\\frac{18}{4}-\\frac{\\sqrt{3} \\mathrm{k}}{2}-\\sqrt{6}=0$
$\\frac{18-2 \\sqrt{3} \\mathrm{k}-4 \\sqrt{6}}{4}=0$
18 \u2013 2\u221a3 k \u2013 4\u221a6 = 0
18 \u2013 4\u221a6 = 2\u221a3k
$\\frac{18-4 \\sqrt{6}}{2 \\sqrt{3}}=\\mathrm{k}$
$\\frac{(18-4 \\sqrt{6}) 2 \\sqrt{3}}{2 \\sqrt{3} \\times 2 \\sqrt{3}}=\\mathrm{k}$
$\\frac{18 \\sqrt{3}-12 \\sqrt{2}}{12}=\\mathrm{k}$
3\u221a3 \u2013 2\u221a2 = k
<\/p>\n\n\n\n

(iv) ax2<\/sup> + bx \u2013 10 = 0
Put $x=-\\frac{\\sqrt{3}}{2}$
$a\\left(-\\frac{2}{5}\\right)^{2}+b\\left(-\\frac{2}{5}\\right)-10=0$
$\\frac{4 a}{25}-\\frac{2 b}{5}-10=0$
$\\frac{4 a-10 b-250}{25}=0$
4a \u2013 10b \u2013 250 = 0
4a \u2013 10b = 250 (1)
Put x = 3\/5
ax2<\/sup> + bx \u2013 10 = 0
$a\\left(\\frac{5}{3}\\right)^{2}+b\\left(\\frac{5}{3}\\right)-10=0$
$\\frac{25 a}{9}+\\frac{5 b}{3}-10=0$
$\\frac{25 a+15 b-90}{9}=0$
25a + 15b \u2013 90 = 0 (divide by 5)
5a + 3b = 18 (2)
Eliminate (1) and (2)
4a \u2013 10b = 250 \u00d75
5a + 3b = 18 \u00d74<\/p>\n\n\n\n

$\\begin{aligned}20a-50b&=1250\\\\-2a-12b&=-72\\\\ \\hline -62b&=1178\\end{aligned}$
<\/p>\n\n\n\n

b = \u201319
Put b = \u201319 in (1)
4a \u2013 10b = 250
4a \u2013 10(\u201319) = 250
4a + 190 = 250
4a = 60
a = 15<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

2x2<\/sup> \u2014 5x + 3 = 0<\/strong>
Sol :
2x\u2013 2x \u2013 3x + 3 = 0
2x (x\u2013 1) \u2013 3(x \u2013 1) = 0
(2x \u2013 3) (x \u2013 1) = 0
2x \u2013 3 = 0
$x=\\frac{3}{2}$
x \u2013 1 = 0
x =1
Therefore, the roots of the equation are $\\frac{3}{2}$, 1.<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

3x2<\/sup> \u20142\u221a6x +2 = 0<\/strong>
Sol :
3x2<\/sup> \u2013 \u221a6 x \u2013 \u221a6 x + 2 = 0
3x2<\/sup> \u2013 \u221a2\u221a3x \u2013 \u221a2\u221a3x + 2 = 0
\u221a3x(\u221a3x \u2013 \u221a2) \u2013 \u221a2 (\u221a3x \u2013 \u221a2) = 0
\u221a3x \u2013 \u221a2 = 0 \u221a3x \u2013 \u221a2 = 0
$\\mathrm{x}=\\frac{\\sqrt{2}}{\\sqrt{3}}, \\mathrm{x}=\\frac{\\sqrt{2}}{\\sqrt{3}}$
Therefore, the roots of the equation are $\\frac{\\sqrt{2}}{\\sqrt{3}}, \\frac{\\sqrt{2}}{\\sqrt{3}}$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

3x2<\/sup> \u2014 14x \u2014 5 = 0<\/strong>
Sol :
3x2<\/sup> \u2014 15x + x \u2014 5 = 0
3x (x \u2013 5) + (x \u2013 5) = 0
(3x + 1) (x \u2013 5) = 0
3x + 1 = 0 x \u2013 5 = 0
$x=-\\frac{1}{3}$, x=5
Therefore, the roots of the equation are $-\\frac{1}{3}, 5$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

$\\sqrt{3} \\mathrm{x}^{2}+10 \\mathrm{x}+7 \\sqrt{3}=0$<\/strong>
Sol :
Now, to find the roots by factorisation, we need to factorise 10 such that the sum is 10 and the product is <\/p>\n\n\n\n

7\u221a3\u00d7\u221a3=21
We can do that by 7 and 3.
So,
\u221a3x2<\/sup> + 10x + 7\u221a3 = 0
\u221a3x2<\/sup> + 3x + 7x + 7\u221a3 = 0
\u221a3x(x + \u221a3) + 7(x + \u221a3) = 0
(\u221a3x + 7)(x + \u221a3) = 0
(\u221a3x + 7) = 0
$\\sqrt{3} x=-7$
$x=-\\frac{7}{\\sqrt{3}}$
Or
x + \u221a3 = 0
x = -\u221a3
Hence, the solutions of the given quadratic equations are -\u221a3 and $-\\frac{7}{\\sqrt{3}}$.<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

\u221a7 y2<\/sup> \u2013 6y \u2014 13\u221a7 = 0<\/strong>
Sol :
\u221a7 y2<\/sup> \u2013 13y + 7y \u2014 13 \u221a7 = 0
y (\u221a7 y \u2013 13) + \u221a7 (\u221a7 y \u2013 13) = 0
(\u221a7 y \u2013 13) (y + \u221a7) = 0
\u221a7 y \u2013 13 = 0 y + \u221a7 = 0
$\\mathrm{y}=\\frac{13}{\\sqrt{7}}$ y = \u2013\u221a7
Rationalise
$\\mathrm{y}=\\frac{13}{\\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$
$\\mathrm{y}=\\frac{13 \\sqrt{7}}{7}$
Therefore, the roots of the equation are $\\frac{13 \\sqrt{7}}{7}-\\sqrt{7}$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations by factorisation:<\/strong><\/p>\n\n\n\n

4x2<\/sup> \u2014 4a2<\/sup>x + a4<\/sup> \u2014 b4<\/sup> = 0<\/strong>
Sol :
4x2<\/sup> \u2013 {2(a2<\/sup> + b2<\/sup>) + 2(a2<\/sup> \u2013 b2<\/sup>)}x + (a2<\/sup> + b2<\/sup>) (a2<\/sup> \u2013 b2<\/sup>) = 0
4x2<\/sup> \u2013 2(a2<\/sup> + b2<\/sup>)x + 2(a2<\/sup> \u2013 b2<\/sup>) x + (a2<\/sup> + b2<\/sup>) (a2<\/sup> \u2013 b2<\/sup>) = 0
2x {2x \u2013 (a2<\/sup> + b2<\/sup>)} \u2013 (a2<\/sup> \u2013 b2<\/sup>) {2x \u2013 (a2<\/sup> + b2<\/sup>)} = 0
{2x \u2013 (a2<\/sup> \u2013 b2<\/sup>)} {2x \u2013 (a2<\/sup> + b2<\/sup>)} = 0
2x \u2013 (a2<\/sup> \u2013 b2<\/sup>) = 0 2x \u2013 (a2<\/sup> + b2<\/sup>) = 0
2x = (a2<\/sup> \u2013 b2<\/sup>) 2x = (a2<\/sup> + b2<\/sup>)
$x=\\frac{a^{2}-b^{2}}{2}, x=\\frac{a^{2}+b^{2}}{2}$
Therefore, the roots of the equation are $\\frac{a^{2}-b^{2}}{2} ,\\frac{a^{2}+b^{2}}{2}$<\/p>\n\n\n\n

\n\n\n\n

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

a2<\/sup>b2<\/sup>x2<\/sup> + b2<\/sup>x \u2013 a2<\/sup>x \u2014 1 = 0, a \u2260 0, b \u2260 0<\/strong><\/p>\n\n\n\n

Sol :
b2<\/sup>x {a2<\/sup>x + 1} \u2013 1 {a2<\/sup>x + 1} = 0
(b2<\/sup>x \u2013 1) (a2<\/sup>x + 1) = 0
b2<\/sup>x \u2013 1 = 0 a2<\/sup>x + 1 = 0
$x=\\frac{1}{b^{2}}, x=\\frac{-1}{a^{2}}$
Therefore, the roots of the equation are $\\frac{1}{b^{2}}, \\frac{-1}{a^{2}}$<\/p>\n\n\n\n

\n\n\n\n

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

36x2<\/sup> \u2014 12ax + (a2<\/sup> \u2014 b2<\/sup>) = 0<\/strong><\/p>\n\n\n\n

Sol :
(6x)2<\/sup> \u2013 2 (6x)a + a2<\/sup> \u2014 b2<\/sup> = 0
Using Identity:
(x\u2013 y)2<\/sup> = x2<\/sup> + y2<\/sup> \u2013 2xy
Here, (6x\u2013a)2<\/sup> = (6x)2<\/sup> \u2013 2 (6x) a + a2<\/sup>
(6x \u2013 a)2<\/sup> \u2013 b2<\/sup> = 0
Using identity:
x2<\/sup> \u2013 y2<\/sup> = (a+ b) (a \u2013 b)
(6x \u2013a + b) (6x \u2013 a \u2013 b) =0
6x = a \u2013 b 6x = a + b
$x=\\frac{a-b}{6},x=\\frac{a+b}{6}$
Therefore, the roots of the equation are$\\frac{a-b}{6}, \\frac{a+b}{6}$<\/p>\n\n\n\n

\n\n\n\n

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

10ax2<\/sup> \u2014 6x + 15ax \u2013 9 = 0, a \u2260 0<\/strong><\/p>\n\n\n\n

Sol :
2x (5ax \u2013 3) + 3 (5ax \u2013 3) = 0
2x+3 =0 5ax \u2013 3 = 0
$x=-\\frac{3}{2}, x=\\frac{3}{50}$
Therefore, roots of the equation are$\\frac{3}{50},-\\frac{3}{2}$<\/p>\n\n\n\n

\n\n\n\n

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

12abx2<\/sup> \u2014 (9a2<\/sup> \u2014 8b2<\/sup>) x \u2014 6ab = 0<\/strong><\/p>\n\n\n\n

Sol :
12abx2<\/sup> \u2014 9a2<\/sup> x\u2014 8b2<\/sup>x \u2014 6ab = 0
3ax (4bx \u2013 3a) + 2b (4bx \u2013 3a) = 0
3ax + 2b =0 4bx \u2013 3a =0
$x=-\\frac{2 b}{3 a}, x=\\frac{3 a}{4 b}$
Therefore, roots of the equation are $-\\frac{2 \\mathrm{b}}{3 \\mathrm{a}}, \\frac{3 \\mathrm{a}}{4 \\mathrm{b}}$<\/p>\n\n\n\n

\n\n\n\n

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

4x2<\/sup> \u2014 2 (a2<\/sup> + b2<\/sup>) x + a2<\/sup>b2<\/sup> = 0<\/strong><\/p>\n\n\n\n

Sol :
4x2<\/sup> \u2014 2a2<\/sup>x+ 2b2<\/sup>x + a2<\/sup>b2<\/sup> = 0
2x (2x \u2013 a2<\/sup>) \u2013 b2<\/sup> (2x \u2013 a2<\/sup>) = 0
2x \u2013 b2<\/sup> = 0 2x \u2013 a2<\/sup> = 0
$x=\\frac{b^{2}}{2} ,x=\\frac{a^{2}}{2}$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations, if they exist by the method of completing the square:<\/strong><\/p>\n\n\n\n

5x2<\/sup> \u2013 6x \u2013 2 = 0<\/strong>
Sol :
5x2<\/sup> \u2013 6x \u2013 2 = 0
Dividing by 5
$x^{2}-\\frac{6 x}{5}-\\frac{2}{5}=0$
We know
(a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>
<\/p>\n\n\n\n

Here, a=x and \u20132ab =$-\\frac{6 x}{5}$
\u20132xb = $-\\frac{6 x}{5}$(\u2235 a =x)
\u20132b = $-\\frac{6}{5}$
$\\mathrm{b}=-\\frac{6}{5 \\times(-2)}$
$\\mathrm{b}=\\frac{3}{5}$
\u2234 Equation becomes<\/p>\n\n\n\n

$x^{2}-\\frac{6 x}{5}-\\frac{2}{5}=0$
Add and subtract $\\left(\\frac{3}{5}\\right)^{2}$
$x^{2}-\\frac{6 x}{5}-\\frac{2}{5}+\\left(\\frac{3}{5}\\right)^{2}-\\left(\\frac{3}{5}\\right)^{2}=0$
$x^{2}-\\frac{6 x}{5}+\\left(\\frac{3}{5}\\right)^{2}-\\frac{2}{5}-\\left(\\frac{3}{5}\\right)^{2}=0$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\left(\\frac{3}{5}\\right)^{2}+\\frac{2}{5}$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\frac{9}{25}+\\frac{2}{5}$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\frac{9+2(5)}{25}$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\frac{9+10}{25}$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\frac{19}{25}$
$\\left(x-\\frac{3}{5}\\right)^{2}=\\frac{(\\sqrt{19})^{2}}{(5)^{2}}$
<\/p>\n\n\n\n

Canceling squares both sides
$\\left(x-\\frac{3}{5}\\right)=\\pm \\frac{\\sqrt{19}}{5}$
<\/p>\n\n\n\n

Solving
$\\left(x-\\frac{3}{5}\\right)=\\frac{\\sqrt{19}}{5}\\left(x-\\frac{3}{5}\\right)=-\\frac{\\sqrt{19}}{5}$

$x=\\frac{\\sqrt{19}}{5}+\\frac{3}{5} x=\\frac{-\\sqrt{19}}{5}+\\frac{3}{5}$
$x=\\frac{\\sqrt{19}+3}{5} ,x=\\frac{-\\sqrt{19}+3}{5}$
So, $\\mathrm{x}=\\frac{\\sqrt{19}+3}{5}$ and $x=\\frac{-\\sqrt{19}+3}{5}$ are the roots of the equation.<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations, if they exist by the method of completing the square:<\/strong><\/p>\n\n\n\n

2x2<\/sup> \u2013 5x + 3 = 0<\/strong>
Sol :
2x2<\/sup> \u2013 5x + 3 = 0
Dividing by 2
$x^{2}-\\frac{5 x}{2}+\\frac{3}{2}=0$
$x^{2}-\\frac{5 x}{2}=-\\frac{3}{2}$
<\/p>\n\n\n\n

Add a coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}-\\frac{5 x}{2}+\\left(\\frac{5}{4}\\right)^{2}=-\\frac{3}{2}+\\left(\\frac{5}{4}\\right)^{2}$
$\\left(x-\\frac{5}{4}\\right)^{2}=-\\frac{3}{2}+\\frac{25}{16}$
$\\left(x-\\frac{5}{4}\\right)^{2}=\\frac{-24+25}{16}$
$\\left(x-\\frac{5}{4}\\right)^{2}=\\frac{1}{16}$
$x-\\frac{5}{4}=\\sqrt{\\frac{1}{16}}$
$x=\\frac{1}{4}+\\frac{5}{4}, x=-\\frac{1}{4}+\\frac{5}{4}$
$x=\\frac{6}{4}=\\frac{3}{2}, x=\\frac{4}{4}=1$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations, if they exist by the method of completing the square:<\/strong><\/p>\n\n\n\n

9x2<\/sup> \u2013 15x + 6 = 0<\/strong>
Sol :
Dividing by 9
$x^{2}-\\frac{15 x}{9}+\\frac{6}{9}=0$
$x^{2}-\\frac{15 x}{9}=-\\frac{2}{3}$
<\/p>\n\n\n\n

Add a coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$\\mathrm{x}^{2}-\\frac{15 \\mathrm{x}}{9}+\\left(\\frac{15}{18}\\right)^{2}=-\\frac{2}{3}+\\left(\\frac{15}{18}\\right)^{2}$
$\\left(x-\\frac{15}{18}\\right)^{2}=-\\frac{2}{3}+\\frac{25}{36}$
$\\left(x-\\frac{15}{18}\\right)^{2}=\\frac{25-24}{36}$
$\\left(x-\\frac{15}{18}\\right)^{2}=\\frac{1}{36}$
$x-\\frac{15}{18}=\\frac{\\sqrt{1}}{\\sqrt{36}}$
$x=\\frac{1}{6}+\\frac{5}{6}$, $x=-\\frac{1}{6}+\\frac{5}{6}$<\/p>\n\n\n\n

$x=\\frac{6}{6}=1, x=\\frac{4}{6}=\\frac{2}{3}$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations, if they exist by the method of completing the square:<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 9x + 18 = 0<\/strong>
Sol :
x2<\/sup> \u2013 9x = \u201318
Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}-9 x+\\left(\\frac{9}{2}\\right)^{2}=-18+\\left(\\frac{9}{2}\\right)^{2}$
$\\left(x-\\frac{9}{2}\\right)^{2}=-18+\\frac{81}{4}$
$\\left(x-\\frac{9}{2}\\right)^{2}=\\frac{-72+81}{4}$
$\\left(x-\\frac{9}{2}\\right)^{2}=\\frac{9}{4}$
$x-\\frac{9}{2}=\\frac{\\sqrt{9}}{\\sqrt{4}}$
$x-\\frac{9}{2}=\\frac{\\pm 3}{2}$
$x=-\\frac{3}{2}+\\frac{9}{2}, x=\\frac{3}{2}+\\frac{9}{2}$
$x=\\frac{6}{2}=3, x=\\frac{12}{2}=6$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of the following quadratic equations, if they exist by the method of completing the square:<\/strong><\/p>\n\n\n\n

2x2<\/sup> + x + 4 = 0<\/strong>
Sol :
$x^{2}+\\frac{x}{2}+2=0$
$x^{2}+\\frac{x}{2}=-2$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}+\\frac{x}{2}+\\left(\\frac{1}{4}\\right)^{2}=-2+\\left(\\frac{1}{4}\\right)^{2}$
$\\left(x+\\frac{1}{4}\\right)^{2}=-2+\\frac{1}{16}$
$\\left(x+\\frac{1}{4}\\right)^{2}=\\frac{-32+1}{16}$
$\\left(x+\\frac{1}{4}\\right)^{2}=\\frac{-31}{16}$
$x+\\frac{1}{4}=\\sqrt{-\\frac{31}{16}}$
Since root cannot be negative
<\/p>\n\n\n\n

Therefore, no real roots exist.<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:<\/strong><\/p>\n\n\n\n

2x2<\/sup> \u2013 5x + 3 = 0<\/strong>
Sol :
$x^{2}-\\frac{5 x}{2}+\\frac{3}{2}=0$
$x^{2}-\\frac{5 x}{2}=-\\frac{3}{2}$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}-\\frac{5 x}{2}+\\left(\\frac{5}{4}\\right)^{2}=-\\frac{3}{2}+\\left(\\frac{5}{4}\\right)^{2}$
$\\left(x-\\frac{5}{4}\\right)^{2}=-\\frac{3}{2}+\\frac{25}{16}$
$\\left(x-\\frac{5}{4}\\right)^{2}=\\frac{-24+25}{16}$
$\\left(x-\\frac{5}{4}\\right)^{2}=\\frac{1}{16}$
$x-\\frac{5}{4}=\\frac{\\sqrt{1}}{\\sqrt{16}}$
$x=\\frac{1+5}{4}, x=\\frac{-1+5}{4}$
$x=\\frac{6}{4}=\\frac{3}{2}, x=\\frac{4}{4}=1$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 6x + 4 = 0<\/strong>
Sol :
x2<\/sup> \u2013 6x = \u20134
Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
x2<\/sup> \u2013 6x + (3)2<\/sup> = \u20134 + (3)2<\/sup>
(x \u2013 3)2<\/sup> = \u20134 + 9
x \u2013 3 = \u221a5
x = \u00b1\u221a5 + 3<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:<\/strong><\/p>\n\n\n\n

\u221a5x2<\/sup> + 9x + 4\u221a5 = 0<\/strong>
Sol :
Divide by \u221a5
$x^{2}+\\frac{9 x}{\\sqrt{5}}=-4$
Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}+\\frac{9 x}{\\sqrt{5}}+\\left(\\frac{9}{2 \\sqrt{5}}\\right)^{2}=-4+\\left(\\frac{9}{2 \\sqrt{5}}\\right)^{2}$
$\\left(x+\\frac{9}{2 \\sqrt{5}}\\right)^{2}=-4+\\frac{81}{20}$
$\\left(x+\\frac{9}{2 \\sqrt{5}}\\right)^{2}=\\frac{-80+81}{20}$
$x+\\frac{9}{2 \\sqrt{5}}=\\frac{\\sqrt{1}}{\\sqrt{20}}$
$\\mathrm{x}=\\frac{1}{2 \\sqrt{5}}-\\frac{9}{2 \\sqrt{5}}, \\mathrm{x}=\\frac{-1}{2 \\sqrt{5}}-\\frac{9}{2 \\sqrt{5}}$
$\\mathrm{x}=-\\frac{8}{2 \\sqrt{5}}=-\\frac{4}{\\sqrt{5}}, \\mathrm{x}=-\\frac{10}{2 \\sqrt{5}}=-\\frac{5}{\\sqrt{5}}$
$x=-\\frac{5}{\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}}=-\\frac{5 \\sqrt{5}}{5}=-\\sqrt{5}$<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:<\/strong><\/p>\n\n\n\n

2x2<\/sup> + \u221a15 x + \u221a2 = 0<\/strong>
Sol :
Divide by 2
$\\mathrm{x}^{2}+\\frac{\\sqrt{15 \\mathrm{x}}}{2}=-\\frac{\\sqrt{2}}{2}$
Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$\\mathrm{x}^{2}+\\frac{\\sqrt{15 \\mathrm{x}}}{2}+\\left(\\frac{\\sqrt{15}}{4}\\right)^{2}=-\\frac{\\sqrt{2}}{2}+\\left(\\frac{\\sqrt{15}}{4}\\right)^{2}$
$\\left(x+\\frac{\\sqrt{15}}{4}\\right)^{2}=-\\frac{\\sqrt{2}}{2}+\\frac{15}{16}$
$\\left(x+\\frac{\\sqrt{15}}{4}\\right)^{2}=\\frac{-8 \\sqrt{2}+15}{16}$
Since root cannot be negative
Therefore, it has no real roots.<\/p>\n\n\n\n

\n\n\n\n

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

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:<\/strong><\/p>\n\n\n\n

x2<\/sup> + x + 3 = 0<\/strong>
Sol :
x2<\/sup> + x = \u20133
Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}+x+\\left(\\frac{1}{2}\\right)^{2}=-3+\\left(\\frac{1}{2}\\right)^{2}$
$\\left(x+\\frac{1}{2}\\right)^{2}=-3+\\frac{1}{4}$
$\\left(x+\\frac{1}{2}\\right)^{2}=\\frac{-12+1}{4}$
$\\left(x+\\frac{1}{2}\\right)^{2}=\\frac{-11}{4}$
$x+\\frac{1}{2}=\\sqrt{-\\frac{11}{4}}$
Since root cannot be negative
Therefore, it has no real roots.<\/p>\n\n\n\n

\n\n\n\n

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

Solve the following equations by the method of completion of a square.<\/strong><\/p>\n\n\n\n

5x2<\/sup> \u2013 24x \u2013 5 = 0<\/strong>
Sol :
Dividing by 5
$x^{2}-\\frac{24 x}{5}-\\frac{5}{5}=0$
$x^{2}-\\frac{24 x}{5}=1$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides<\/p>\n\n\n\n

$x^{2}-\\frac{24 x}{5}+\\left(\\frac{24}{10}\\right)^{2}=1+\\left(\\frac{24}{10}\\right)^{2}$
$\\left(x-\\frac{24}{10}\\right)^{2}=1+5.76$
$x-\\frac{24}{10}=\\sqrt{6.76}$
x = 2.6 + 2.4 x = \u20132.6 + 2.4
x = 5 x = \u2013 0.2<\/p>\n\n\n\n

\n\n\n\n

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

Solve the following equations by the method of completion of a square.<\/strong><\/p>\n\n\n\n

7x2<\/sup> \u2013 13x \u2013 2 = 0<\/strong>
Sol :
Divide by 7
$x^{2}-\\frac{13 x}{7}=\\frac{2}{7}$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}-\\frac{13 x}{7}+\\left(\\frac{13}{14}\\right)^{2}=\\frac{2}{7}+\\left(\\frac{13}{14}\\right)^{2}$
$\\left(x-\\frac{13}{14}\\right)^{2}=\\frac{2}{7}+\\frac{169}{196}$
$\\left(x-\\frac{13}{14}\\right)^{2}=\\frac{56+169}{196}$
$\\left(x-\\frac{13}{14}\\right)^{2}=\\frac{225}{196}$
$x-\\frac{13}{14}=\\frac{\\sqrt{225}}{\\sqrt{196}}$
$x-\\frac{13}{14}=\\frac{15}{14}$
$x=\\frac{15+13}{14} ,x=\\frac{-15+13}{14}$
$x=\\frac{28}{14}=2 x=-\\frac{2}{14}=-\\frac{1}{7}$<\/p>\n\n\n\n

\n\n\n\n

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

Solve the following equations by the method of completion of a square.<\/strong><\/p>\n\n\n\n

15x2<\/sup> + 53x + 42 = 0<\/strong>
Sol :
Divide by 15
$x^{2}+\\frac{53 x}{15}=-\\frac{42}{15}$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}+\\frac{53 x}{15}+\\left(\\frac{53}{30}\\right)^{2}=-\\frac{42}{15}+\\left(\\frac{53}{30}\\right)^{2}$
$\\left(x+\\frac{53}{30}\\right)^{2}=-\\frac{42}{15}+\\frac{2809}{900}$
$\\left(x+\\frac{53}{30}\\right)^{2}=\\frac{-2520+2809}{900}$
$x+\\frac{53}{30}=\\frac{\\sqrt{289}}{\\sqrt{900}}$
$x=\\frac{17}{30}-\\frac{53}{30}, x=-\\frac{17}{30}-\\frac{53}{30}$
$x=-\\frac{6}{5} ,x=-\\frac{7}{3}$<\/p>\n\n\n\n

\n\n\n\n

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

Solve the following equations by the method of completion of a square.<\/strong><\/p>\n\n\n\n

7x2<\/sup> + 2x \u2013 5 = 0<\/strong>
Sol :
Divide by 7
$x^{2}+\\frac{2 x}{7}=\\frac{5}{7}$
<\/p>\n\n\n\n

Add the coefficient of $\\left(\\frac{x}{2}\\right)^{2}$ to both sides
$x^{2}+\\frac{2 x}{7}+\\left(\\frac{2}{14}\\right)^{2}=\\frac{5}{7}+\\left(\\frac{2}{14}\\right)^{2}$
$\\left(x+\\frac{2}{14}\\right)^{2}=\\frac{5}{7}+\\frac{4}{196}$
$\\left(x+\\frac{2}{14}\\right)^{2}=\\frac{140+4}{196}$
$\\left(x+\\frac{2}{14}\\right)^{2}=\\frac{144}{196}$
$x+\\frac{2}{14}=\\frac{\\sqrt{144}}{\\sqrt{196}}$
$x+\\frac{2}{14}=\\frac{12}{14}$
$x=\\frac{12-2}{14} ,x=\\frac{-12-2}{14}$
$x=\\frac{10}{14}=\\frac{5}{7} ,x=-\\frac{14}{14}=-1$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 Determine whether $x=\\frac{3}{2}$ and $x=-\\frac{4}{3}$ are the solutions of the equation 6×2-x-12=0 or not. Sol :Put both the values of x in the equation.When $x=\\frac{3}{2}$$6\\left(\\frac{3}{2}\\right)^{2}-\\frac{3}{2}-12=0$$6 \\times \\frac{9}{4}-\\frac{3}{2}-12=0$$\\frac{54-48-6}{4}$$\\frac{54-54}{4}$= 0 When $x=-\\frac{4}{3}$$6\\left(-\\frac{4}{3}\\right)^{2}-\\frac{4}{3}-12=0$$6\\left(\\frac{16}{9}\\right)+\\frac{4}{3}-12=0$$\\frac{96+12-108}{9}$$\\frac{108-108}{9}$= 0R.H.S = L.H.STherefore, $\\mathrm{x}=\\frac{3}{2}$ and $x=-\\frac{4}{3}$ are the solutions of the given equation. Question 2  Determine whether (i) x = 1, (ii) x = 3 are the solutions of the equation x2 \u2014 5x […]<\/p>\n","protected":false},"author":302,"featured_media":624299,"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":[24],"tags":[],"boards":[],"class_list":["post-624304","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 7.3 - Mathematics Solution Class 10 Chapter 7 Quadratic Equations - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 Determine whether $x=frac{3}{2}$ and $x=-frac{4}{3}$ are the solutions of the equation 6x2-x-12=0 or not. Sol :Put both\" \/>\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-7-3-mathematics-solution-class-10-chapter-7-quadratic-equations\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 7.3 - Mathematics Solution Class 10 Chapter 7 Quadratic Equations\" \/>\n<meta property=\"og:description\" content=\"Question 1 Determine whether $x=frac{3}{2}$ and $x=-frac{4}{3}$ are the solutions of the equation 6x2-x-12=0 or not. 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