Question

what are the real or imaginary solutions of the polynomial equation x^4-41x^2=-400

Answer 1

Let u=x²
Then solve:
u²-41u+400 = 0
The x values will be ±sqrt(u)

Answer 2

this is an equation that can be reduced to a quadratic equation

\[x ^{4}-41x ^{2}+400=0\]


so you have factors of 400 that add to -41 are -25 and - 16

so you need to solve

\[(x ^{2}−16)(x ^{2}−25)=0\]


each quadratic in the brackets above are the difference of 2 squares. Hope this helps

Answer 3

what do you mean about the difference of 2squares

Answer 4

The difference of two squares rule:
(a²-b²) = (a-b)(a+b)

Since 16 = 4², we can say:
(x² - 16) = (x+4)(x-4)