Question

Question #301c5

Answer 1

I would say no. Let us solve the equation for `x`.

`|x^2-a|=|x^2+b|`

By removing the absolute value sign,

Case 1
`x^2-a=x^2+b Rightarrow -a=b`

Case 2
`x^2-a=-(x^2+b)`

by adding `x^2` and `a`,

`Rightarrow2x^2=a-b`

by dividing by 2,

`Rightarrow x^2={a-b}/2`

by taking the square-root,

`Rightarrow x=pmsqrt{{a-b}/2}`

Hence, if `b=-a`, then the solution is any real number; otherwise, `x=pmsqrt{{a-b}/2}`.

Note
I am guessing that whoever came up with this question was thinking to plug `b=-a` (Case 1) into `x^2={a-b}/2` (Case 2); however, that is incorrect.

---

I hope that this was helpful.