# Question #301c5

Oct 21, 2014

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
${x}^{2} - a = {x}^{2} + b R i g h t a r r o w - a = b$

Case 2
${x}^{2} - a = - \left({x}^{2} + b\right)$

by adding ${x}^{2}$ and $a$,

$R i g h t a r r o w 2 {x}^{2} = a - b$

by dividing by 2,

$R i g h t a r r o w {x}^{2} = \frac{a - b}{2}$

by taking the square-root,

$R i g h t a r r o w x = \pm \sqrt{\frac{a - b}{2}}$

Hence, if $b = - a$, then the solution is any real number; otherwise, $x = \pm \sqrt{\frac{a - b}{2}}$.

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

I hope that this was helpful.