What are the real or imaginary solutions of the polynomial equation?
x^4 – 41x^2 = –400

Question

What are the real or imaginary solutions of the polynomial equation?
x^4 – 41x^2 = –400

anonymous · Answer

I need to show the work, I already know the answer is 4, -4, 5, -5

anonymous · Answer

Substitute $u=x^2$ to yield the quadratic $u^2-41u+400=u^2-16u-25u+400=u(u-16)-25(u-16)=(u-25)(u-16)$. If this product equals $0$, then $u\in\{16,25\}$.
If $u=16$, then we back-substitute $u=x^2$ yielding $x^2=16$, i.e. $x=\pm4$.
Similarly, $x^2=25$ and thus $x=\pm5$.

anonymous · Answer

Hmm.. Most of that doesn't look familiar to me... this is Algebra 2

anonymous · Answer

Or wait, I guess I am just confused about the "u" thing. Why substitute that in and not just use x?

anonymous · Answer

If we let $u=x^2$, we can rewrite the equation as $u^2 - 41u = -400$, which is then equivalent to $u^2-41u+400=0$. If you're in Algebra II you should understand how to factor this... it yields $(u-16)(u-25)=0$. If the product of these two factors is $0$, then it follows logically that either one of them must be $0$ -- so:
$$u-16=0\u=16\\lor\u-25=0\u=25$$By substituting back $u=x^2$, we now have:$$x^2=16\x=\pm\sqrt{16}=\pm4\\lor\x^2=25\x=\pm\sqrt{25}=\pm5$$

anonymous · Answer

We substitute so that it's more readily visible as a quadratic of u, which we know how to solve by factoring.

anonymous · Answer

OHHHhhhh I see, A million thanks for your help! :)

anonymous · Answer

oh and is "u" like commonly used in this type of case or was it just your choice of letter?