Question

Find the value of $k$, if the following equations have equal real roots:
$(k+1)x^2-2(k+4)x+2k=0$, $k\in R$

Answer 1

Given:

The given equation is $(k+1)x^2-2(k+4)x+2k=0$

On comparing the given equation with standard quadratic equation $a{x}^2+bx+c=0$, we have

$a=k+1,b=-2k-8,c=2k$

It is given that the equation has real equal roots, so its Discriminant, $D=0$

Discriminant $=b^2-4ac=0$

$\Rightarrow {(-2k-8)}^2-4\times (k+1)\left(2k\right)=0$

$\Rightarrow {4(k+4)}^2-4\times (k+1)\left(2k\right)=0$ [Taking $(-2)$ outside with square]

$\Rightarrow {(k+4)}^2-2k(k+1)=0$ [on dividing both sides by $4$]

$\Rightarrow k^2+8k+16-2k^2-2k=0$ [Using the identity, $(a+b{)}^2=a^2+2ab+b^2$]

$\Rightarrow -k^2+6k+16=0$

$\Rightarrow k^2-6k-16=0$

$\Rightarrow k^2-8k+2k-16=0$

$\Rightarrow k(k-8)+2(k-8)=0$

$\Rightarrow (k-8)(k+2)=0$

$\Rightarrow (k-8)=0or,(k+2)=0$

$\therefore k=8,-2$