Question

For each equation, given below, find the value of ′m′ so that equation has equal roots. Also, find the solution of each equation.

(i). $(m-3)x^2-4x+1=0$

Answer 1

Given equation: $(m-3)x^2-4x+1=0$

For equal roots, Discriminant $\left(D\right)=0$ where $D=b^2-4ac$

Comparing given equation $(m-3)x^2-4x+1=0$ with $a{x}^2+bx+c=0$

We have $a=m-3,b=-4$ and $c=1$

$D=(-4{)}^2-4(m-3)\times 1=0$

$\{\because D=b^2-4ac\}$

$\Rightarrow 16-4m+12=0$

$\Rightarrow 4m=28$

$\Rightarrow m=\frac{28}{4}=7$

Put value of m in given equation we get

$4x^2-4x+1=0$

$\Rightarrow (2x-1{)}^2=0$ $\{\because a^2-2ab+b^2=(a-b{)}^2\}$

$\Rightarrow 2x-1=0$

$\Rightarrow x=\frac{1}{2}$

$\therefore m=7$ and $x=\frac{1}{2}$.