Logan Wyatt

2022-07-08

Given that $a$, $b$ and $c$ are positive real numbers that satisfy
$b=\frac{64a}{{a}^{2}-64}=\frac{81c}{2{c}^{2}-81}=\sqrt{{a}^{2}+{c}^{2}}$, find $b$.

vrtuljakwb

Expert

Squaring yields two polynomial equations in $a$ and $c$,
${a}^{6}+{a}^{4}{c}^{2}-128{a}^{4}-128{a}^{2}{c}^{2}+4096{c}^{2}=0,$
and
$-81{a}^{2}c+128a{c}^{2}-5184a+5184c=0.$
Over the complex numbers all solutions can be computed by using Groebner bases. Among them the positive real solutions are $\left(a,b,c\right)=\left(0,0,0\right)$, $\left(a,b,c\right)=\left(24/\sqrt{5},30/\sqrt{5},18/\sqrt{5}\right)$. All other solutions are either real with one of the values $a,b,c$ negative, or non-real solutions. We have ${b}^{2}={a}^{2}+{c}^{2}$, which is, up to a factor here ${30}^{2}={24}^{2}+{18}^{2}$, i.e., ${5}^{2}={4}^{2}+{3}^{2}$.

Jamison Rios

Expert