Question

The value of $k$ for which the quadratic equation $k{x}^2+1=kx+3x-11x^2$ has real and equal roots are

Answer 1

We have, $k{x}^2+1=kx+3x-11x^2$

$\Rightarrow (k+11)x^2-(k+3)x+1=0$

On comparing with general form of Quadratic Equation $a{x}^2+bx+c=0$

We get $a=(k+11),b=-(k+3),c=1$

Since the quadratic equation has real and equal roots

$D=0$

$\Rightarrow b^2-4ac=0$

$\Rightarrow (-(k+3){)}^2-4(k+11)\times 1=0$

$\Rightarrow (k^2+6k+9)-4k-44=0$

$\Rightarrow k^2+2k-35=0$

$\Rightarrow k^2+7k-5k-35=0$

$\Rightarrow k(k+7)-5(k+7)=0$

$\Rightarrow (k+7)(k-5)=0$

$\Rightarrow (k+7)=0$ or $(k-5)=0$

$\Rightarrow k=-7$ or $k=5$

Hence, Option (C) is correct.