For how many integers $a$ does this equation have three solutions?

Question

For how many integers $a$ does the equation $(x^2-a^2 ) \sqrt{(5-x)}=0$ have three different solutions?

The options were: $10, 9, 8, $other.

I say other.

No matter what, $\sqrt{5-x} = 0$ always has one solutions.

Then it is remaining to solve:

$x^2 - a^2 = 0 \implies x^2 = a^2 \implies x = \pm a$, which is true for any $a$ really, except $a=0$.

Answer 1

Assuming:

1. Solutions should be integers
2. Imaginary numbers do not come to picture.

We need three distinct solutions.

Consider $(x^2-a^2)=0\implies x=\pm a$ if $a\ne0$, else two solutions will be $0$.

Consider $\sqrt{5-x}=0\implies x\le5$, else $\sqrt{5-x}$ goes imaginary.

If $a=5$, then two solutions will be $5$, one contributed by $\sqrt{5-x}=0$ and one by $(x^2-5^2)=0$.

Hence the only integer options left with us are $\pm 1,\pm2,\pm3,\pm4$, that is, 8 choices for $a$.

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However, if imaginary numbers are allowed, all values of $a$ except $0,5$ works.

Answer 2

For $\sqrt{5-x}$ to be real $x\le5$

So values x can hold are 1,2,3,4,5 For $(x^2-a^2) = 0 $ x should be $\pm a$ But if it becomes $\pm 5$ there will be only two solutions because one 5 will be contributed from $\sqrt{5-x} = 0 $ hence the required numbers are $\pm 1, \pm 2 , \pm3 , \pm4 $