**Exercise 8.1**

**1. (i) x ^{2Â }– 3x + 2 = 0 , x = 2 , x = -1**

Here LHS = x^{2}-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,Â 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 â‰ RHS

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

**(ii) x ^{2}+ x+1 =0 , x= 0 , x= 1**

Here, LHS = x^{2 }+x+1 and RHS =0

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

= 0^{2}+0+1Â = (1)^{2}+1+1 = 1 =3

LHS â‰ 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 =Â Â \(x^{2}-3\sqrt{3}x+6=0\)

Substituting the value of \(x=\sqrt{3}\,and\,x=-2\sqrt{3}\)

\(\sqrt{3}^{2}-3\sqrt{3}\times \sqrt{3}+6\)

= 3-9+6

= 0

= RHS

\(-2\sqrt{3}^{2}-3\sqrt{3}\times -2\sqrt{3}+6\)

= 12+18+6

= 36

â‰ RHS

\(x=\sqrt{3}\)

Whereas, \(x=-2\sqrt{3}\)

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

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

Here, LHS = \(x+\frac{1}{x}=\frac{13}{6}\)

Substituting \(where\,x=\frac{5}{6}\,and\,x=\frac{4}{3}\)

= \(\frac{5}{6}+\frac{1}{\frac{5}{6}}\)

= \(\frac{5}{6}+\frac{6}{5}\)

= \(\frac{25+36}{30}\)

= \(\frac{61}{30}\)

â‰ RHS

= \(\frac{4}{3}+\frac{1}{\frac{4}{3}}\)

= \(\frac{4}{3}+\frac{3}{4}\)

= \(\frac{16+9}{12}\)

=\(\frac{25}{12}\)

â‰ RHS

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

**(v)** 2x^{2}-x+9 = x^{2}+4x+3 , x = 2 and x = 3

= 2x^{2}-x+9 – x^{2}+4x+3

= x^{2} -5x+6 = 0

Here, LHS = x^{2} -5x+6 and RHS = 0

Substituting x = 2 and x = 3

= x^{2} -5x+6

= (2)^{2} â€“ 5(2) +6

=10-10

=0

= RHS

= x^{2} -5x+6

= (3)^{2} â€“ 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 \(x=-\sqrt{2}\,and\,x=-2\sqrt{2}\)

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

= 2 â€“ 2 – 4

= -4

â‰ RHS

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

= 8 -4 -4

=8-8

=0

= RHS

\(x=-2\sqrt{2}\)

**(vii) Â **a^{2}x^{2}-3abx+2b^{2} = 0

\(x=\frac{a}{b}\,and\,x=\frac{b}{a}\)

Here, LHS = a^{2}x^{2} -3abx+2b^{2 }and RHS = 0

Substituting the \(x=\frac{a}{b}\,and\,x=\frac{b}{a}\)

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

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

â‰ RHS

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

= b^{2}-3b^{2} +2b^{2Â }= 0 = RHS

\(x=\frac{b}{a}\)

**3.**

**(i)** Given that \(\frac{2}{3}\)

The equation is 7x^{2}+kx-3 = 0

According to the question \(\frac{2}{3}\)

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

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

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

= \(\frac{2k}{3}=\frac{-1}{9}\)

= \(k=\frac{-1}{6}\)

**(ii)** Given that x=a is a root of the given equation x^{2}-x(a+b)+k =0

= x=a satisfies the equation

= a^{2}-a(a+b)+k=0

= a^{2} – a^{2}-ab+k =0

K = ab

**(iii)** Given that \(x=\sqrt{2}\)

\(x=\sqrt{2}\)

=\(k\sqrt{2}^{2}+\sqrt{2}\sqrt{2}-4\)

=2k+2-4 =0

= 2k-2=0

K = 1

**(iv)** Given that x= -a is the root of the given equation x^{2}+3ax+k = 0

Therefore,

= (-a)^{2}+3a(-a)+k = 0

= a^{2}+3a^{2}+k =0

= k = 4a^{2}= -a satisfies the equation

**(v)** Given that \(x=\sqrt{2}\)

\(x=\sqrt{2}\)

=\(k\sqrt{2}^{2}+\sqrt{2}\sqrt{2}-4\)

=2k+2-4 =0

= 2k-2=0

K = 1

**4.Â **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

\(\sqrt{3^{2}-4\times3+3}\)

\(\sqrt{9-12+3}\)

\(\sqrt{12-12}\)

= 0

Similarly putting x=3 in RHS

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

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

\(\sqrt{52-42}\)

\(\sqrt{10}\)

â‰ RHS

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