Question

solve x^4-15*x^2+56=0

Answer 1

let y = x^2, then we have:
\[y = x^2 \Rightarrow y^2 = x^4, x^4-15x^2+56 = 0 \iff y^2-15y+56 = 0\]
We can factor that equation of y by thinking of two numbers that multiply to 56, but add to -15. Those numbers are -8 and -7:
\[y^2-15y+56 \iff (y-8)(y-7) = 0\]
Replacing y = x^2 gives:
\[(x^2-8)(x^2-7) = 0\]
So the four solutions are:
\[x^2-7 = 0 \iff x = \pm \sqrt{7}, x^2-8 = 0 \iff x = \pm \sqrt{8}\]

Answer 2

Y^2+15Y+56=0