Question

Solve: $b^4+3b^2-40$

Answer 1

$b^4+3b^2-40$

taking $b^2=t$

$t^2+3t-40$

for roots - $t^2+3t-40=0$

$t=\frac{-3\pm \sqrt{(3{)}^2-4\times (-40)\times 1}}{2\times 1}$

$t=\frac{-3\pm \sqrt{169}}{2}=\frac{-3\pm 13}{2}$

$t=\frac{-16}{2},\frac{10}{2}=-8,5$

$b^2=-8$ and $b^2=5$

roots of the equation $\to b=\pm \sqrt{5}$