Question

Solution set for (x^2-6x)^2-35(x^2-6x)-200=0

Answer 1

let u=x^2-6x

then u^2=(x^2-6x)^2

so we have
u^2-35u-200=0

\[u=\frac{35 \pm \sqrt{(-35)^2-4(1)(-200)}}{2(1)}=\frac{35 \pm \sqrt{1225+800}}{2}=\frac{35 \pm \sqrt{2025}}{2}\]
\[=\frac{35 \pm 45}{2}=\frac{35+45}{2},\frac{35-45}{2}=\frac{80}{2}.\frac{-10}{2}=40,-5\]

but remember u=x^2-6x
so
x^2-6x=40 or x^2-6x=-5
solve both of these equations for x