Find $a$ and $b$, where $b0$, such that $$|x^2-a|=b$$ has exactly the solutions \[-4,

Question

Find $a$ and $b$, where $b>0$, such that $$|x^2-a|=b$$ has exactly the solutions $$-4,-2,2,4.$$Then generalise to the four solutions $\pm p,\pm q$, where $0<p<q$.

Accepted Answer

$a=10,b=6$; generally $a=\frac{p^2+q^2}{2}$, $b=\frac{q^2-p^2}{2}$.

Modulus Function

Modulus Function

Solution:

Solution locked