Question

find the value of k for which the following equation has two equal roots.

(k+2)x^2+4k=(4k+2)x

Answer 1

Compute the discriminant of the quadratic $a x^{2}+b x+c=0.$

It is:

$\Delta=b^{2}-4 a c=(4 k+2)^{2}-4 \cdot 4 k(k+2)\\=16 k^{2}+16 k+4-16 k^{2}-32 k=-16 k+4$

A quadratic has two equal roots if and only if $\Delta=0\\$

$-16k+4=0\Rightarrow k=\frac{4}{16}=\frac{1}{4}$

Hence, the value of k is $\frac{1}{4}$ .