Question

For what value of $k$, the equation $x^2+2(k-4)x+2k=0$ has equal roots ?

Answer 1

Step 1: Estimate the constants of the equation

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

For equal roots,

$b^2-4ac=0$

In this case $a=1,b=2\left(k-4\right),c=2k$.

Step 2: Calculate the value of $k$

Hence,

$b^2-4ac=0$

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

$\Rightarrow$$4k^2-32k+64-8k=0$

$\Rightarrow$ $4k^2-40k+64=0$

$\Rightarrow$ $k^2-10k+16=0$

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

$\Rightarrow$ $k\left(k-8\right)-2\left(k-8\right)=0$

$\Rightarrow$ $\left(k-8\right)\left(k-2\right)=0$

$\Rightarrow$ $k=2,8$

Hence, the required values of $k$ are $2$ and $8$.