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?

Your approach is correct (even if things could have been done faster just as Ahmed S. Attaalla answered) but the mistake is $\Delta=16-4(4-2a)=8a$ and the roots are given by $x_{1,2}=\frac{4\pm\sqrt{8a}}2=2\pm \sqrt{2a}$ — Claude Leibovici, Feb 23 '17 at 05:31

score 2 · Accepted Answer · answered Feb 23 '17 at 02:01

2

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 $2\pm \sqrt{2a}$.

answered Feb 23 '17 at 02:01

Ahmed S. Attaalla

1 Answers1