Value of $a$ for $x^{2}-4x+4-2a=0$ to have a unique solution

Question

By using the root formula i got $(2\pm \sqrt{8a})$ expression.

So, $ a $ cannot be negative and if $a$ is positive we get two solutions.

Therefore, $a=0$ is the answer??

Is my approach correct and is there a better way to solve this problem?

Answer 1

Yes ,your approach is correct but is seems you incorrectly calculated the roots. It is enough to note that we seek a unique solution to

$$(x-2)^2=2a$$

This is only possible if $a=0$.

And the corresponding solutions are $2\pm \sqrt{2a}$.