Question

Find the value of $k$, if the equation $4x^2-2(k+1)x+(k+1)=0$ has real roots and equal roots.

Answer 1

Given: Equation $4x^2-2(k+1)x+(k+1)=0$ has real and equal roots i.e., its discriminant is zero.

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

For given equation, $a=4,b=2(k+1),c=(k+1)$

$\therefore \left(2\right(k+1){)}^2-4.4(k+1)=0$

$\Rightarrow (2k+2{)}^2-16(k+1)=0$

$\Rightarrow 4k^2+8k+4-16k-16=0$

$\Rightarrow 4k^2-8k-12=0$

$\Rightarrow 4(k^2-2k-3)=0$

$\Rightarrow k^2-2k-3=0$

$\Rightarrow k^2-3k+k-3=0$

$\Rightarrow k(k-3)+1(k-3)=0$

$\Rightarrow (k-3)(k+1)=0$

$\Rightarrow k-3=0$ and $k+1=0$

$\Rightarrow k=3$ and $k=-1$

$\therefore$ The values of $k$ are $3$ and $-1.$