# RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.1

## RD Sharma Solutions Class 10 Chapter 8 Exercise 8.1

### RD Sharma Class 10 Solutions Chapter 8 Ex 8.1 PDF Free Download

#### Exercise 8.1

1. (i) x– 3x + 2 = 0 , x = 2 , x = -1

Here LHS = x2-3x+2

RHS = 0

Now, substitute x = 2 in LHS

We get,

(2)2 -3(2)+2 = 4-6+2 = 6-6 =0 = RHS

Since, LHS =RHS

Therefore,  x – 2 is a solution of the given equation.

Similarly, substituting x= -1 in LHS

We get,

(-1)2 -3(-1)+2 = 1+3+2 = 6 ≠ RHS

Since, LHS ≠ RHS = x= -1 is not the solution of the given equation.

(ii) x2+ x+1 =0 , x= 0 , x= 1

Here, LHS = x2 +x+1 and RHS =0

Now, substituting x= 0 and x= 1 in LHS

= 02+0+1  = (1)2+1+1 = 1 =3

LHS ≠ RHS

Both x=0 and x=1 are not solutions of the given equation.

(iii) $x^{2}-3\sqrt{3}x+6=0$ , $x=\sqrt{3}\,and\,x=-2\sqrt{3}$

Here,

LHS =   $x^{2}-3\sqrt{3}x+6=0$ and RHS =0

Substituting the value of $x=\sqrt{3}\,and\,x=-2\sqrt{3}$ in LHS

$\sqrt{3}^{2}-3\sqrt{3}\times \sqrt{3}+6$

= 3-9+6

= 0

= RHS

$-2\sqrt{3}^{2}-3\sqrt{3}\times -2\sqrt{3}+6$

= 12+18+6

= 36

≠ RHS

$x=\sqrt{3}$ is a solution of the above mentioned equation

Whereas, $x=-2\sqrt{3}$ is not a solution of the above mentioned equation .

(iv) $x+\frac{1}{x}=\frac{13}{6}$

$where\,x=\frac{5}{6}\,and\,x=\frac{4}{3}$

Here, LHS = $x+\frac{1}{x}=\frac{13}{6}$    and RHS = $\frac{13}{6}$

Substituting $where\,x=\frac{5}{6}\,and\,x=\frac{4}{3}$ in the LHS

= $\frac{5}{6}+\frac{1}{\frac{5}{6}}$

= $\frac{5}{6}+\frac{6}{5}$

= $\frac{25+36}{30}$

= $\frac{61}{30}$

≠ RHS

= $\frac{4}{3}+\frac{1}{\frac{4}{3}}$

= $\frac{4}{3}+\frac{3}{4}$

= $\frac{16+9}{12}$

=$\frac{25}{12}$

≠RHS

$where\,x=\frac{5}{6}\,and\,x=\frac{4}{3}$  are not the solutions of the given equation.

(v) 2x2-x+9 = x2+4x+3 , x = 2 and x = 3

= 2x2-x+9 – x2+4x+3

= x2 -5x+6 = 0

Here, LHS = x2 -5x+6 and RHS = 0

Substituting x = 2 and x = 3

= x2 -5x+6

= (2)2 – 5(2) +6

=10-10

=0

= RHS

= x2 -5x+6

= (3)2 – 5(3) +6

= 9-15+6

=15-15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.

(vi) $x^{2}-\sqrt{2}x-4=0$

$x=-\sqrt{2}\,and\,x=-2\sqrt{2}$

Here, LHS = $x^{2}-\sqrt{2}x-4=0$

And RHS = 0

Substituting the value $x=-\sqrt{2}\,and\,x=-2\sqrt{2}$ in LHS

= $(-\sqrt{2})^{2}-\sqrt{2}\times \sqrt{2}-4$

= 2 – 2 – 4

= -4

≠RHS

= $(-2\sqrt{2})^{2}-\sqrt{2}\times 2\sqrt{2}-4$

= 8 -4 -4

=8-8

=0

= RHS

$x=-2\sqrt{2}$ is the solution of the above mentioned quadratic equation .

(vii)  a2x2-3abx+2b2 = 0

$x=\frac{a}{b}\,and\,x=\frac{b}{a}$

Here, LHS = a2x2 -3abx+2b2 and RHS = 0

Substituting the $x=\frac{a}{b}\,and\,x=\frac{b}{a}$ in LHS

= $a^{2}(\frac{a}{b})^{2}-3ab(\frac{a}{b})+2b^{2}$

= $\frac{a^{4}}{b^{2}}-3a^{2}+2b^{2}$

≠ RHS

= $a^{2}(\frac{b}{a})^{2}-3ab(\frac{b}{a})+2b^{2}$

= b2-3b2 +2b= 0 = RHS

$x=\frac{b}{a}$  is the solution of the above mentioned quadratic equation .

3.

(i) Given that $\frac{2}{3}$ is a root of the given equation.

The equation is 7x2+kx-3 = 0

According to the question $\frac{2}{3}$ satisfies the equation.

= $7(\frac{2}{3})^{2}+k(\frac{2}{3})-3$

= $7(\frac{4}{9})+\frac{2k}{3}-3$

= $\frac{2k}{3}=\frac{27-28}{9}$

= $\frac{2k}{3}=\frac{-1}{9}$

= $k=\frac{-1}{6}$

(ii) Given that x=a is a root of the given equation x2-x(a+b)+k =0

= x=a satisfies the equation

= a2-a(a+b)+k=0

= a2 – a2-ab+k =0

K = ab

(iii) Given that $x=\sqrt{2}$ is a root of the given equation $kx^{2}+\sqrt{2}x-4$

$x=\sqrt{2}$ satisfies the given quadratic equation.

=$k\sqrt{2}^{2}+\sqrt{2}\sqrt{2}-4$

=2k+2-4 =0

= 2k-2=0

K = 1

(iv) Given that x= -a is the root of the given equation x2+3ax+k = 0

Therefore,

= (-a)2+3a(-a)+k = 0

= a2+3a2+k =0

= k = 4a2= -a satisfies the equation

(v) Given that $x=\sqrt{2}$ is a root of the given equation $kx^{2}+\sqrt{2}x-4$

$x=\sqrt{2}$ satisfies the given quadratic equation.

=$k\sqrt{2}^{2}+\sqrt{2}\sqrt{2}-4$

=2k+2-4 =0

= 2k-2=0

K = 1

4. Given to check whether 3 is a root of the equation ${\sqrt{x^{2}-4x+3}}+\sqrt{x^{2}-9}=\sqrt{4x^{2}-14x+16}$

LHS = ${\sqrt{x^{2}-4x+3}}+\sqrt{x^{2}-9}$

RHS= $\sqrt{4x^{2}-14x+16}$

Substituting x=3 in LHS

$\sqrt{3^{2}-4\times3+3}$ + $\sqrt{3^{2}-9}$

$\sqrt{9-12+3}$ + $\sqrt{9-9}$

$\sqrt{12-12}$ + $\sqrt{9-9}$

= 0

Similarly putting x=3 in RHS

${\sqrt{4x^{2}-14x+16}$

$\sqrt{4(3)^{2}-14(3)+16}$

$\sqrt{52-42}$

$\sqrt{10}$

≠ RHS

X=3 is not the solution the given quadratic equation.

#### Practise This Question

State true or false.
Square root of a rational number is also a rational number.