Question 58382
Depending on your math status, I could lead you onto what your professor was indicating.
y = x^2
leads to ~>
(x^2)^2 - 5(x^2) - 8 = 0
(y)^2 - 5(y) - 8 = 0
y^2 - 5y - 8 = 0
{{{y = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{y = (5 +- sqrt( 25 + 32 ))/(2) }}}
{{{y = (5 +- sqrt( 57 ))/(2) }}}
Now, we have the partial answer. Since the original equation supported {{{x}}}, we must have the variable {{{x}}} equal something to develop the answer.
{{{sqrt(y) = x}}}
{{{x = sqrt((5 +- sqrt( 57 ))/(2))}}}
About:
x = 2.5050
x = -2.5050
{{{graph(300,300,-10,10,-10,10,x^4 - 5x^2 - 8)}}}