Question

Find the values of $k$ for which the given equation has real and equal roots
$x^2+k(4x+k-1)+2=0$

Answer 1

The given equation is,

$x^2+k(4x+k-1)+2=0\Rightarrow x^2+4kx+k(k-1)+2=0$

Here, $a=1,b=4k,c=k(k-1)+2=0$

$\therefore D=b^2-4ac={\left(4k\right)}^2-4\times 1\left[k\right(k-1)+2]=16k^2-4k(k-1)-8$

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

$\Rightarrow D=12k^2+4k-8\Rightarrow D=4(3k^2+k-2)$

$\Rightarrow D=4\left[3k\right(k+1)-2(k+1\left)\right]\Rightarrow D=4(3k-2)(k+1)$

The given equation will have equal roots, if

$D=0\Rightarrow 4(3k-2)(k+1)=0\Rightarrow k=\frac{2}{3}ork=-1$