Question

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

Answer 1

http://openstudy.com/study#/updates/5070ece6e4b0c2dc8340b751

Answer 2

Here is a hint:
Your equation is \(x^4 - 41x^2=-400\)
Make a variable, say \(b\), and let it equal to \(x^2\)
So: \(b=x^2\)

In your equation, replace \(x^2\) and \(x^4\) with \(b\)
\(x^4 - 41x^2=-400\)
\(b^2 - 41b=-400\)
Move the \(-400\) to the RHS
\(b^2 - 41b+400=0\)

Now solve for \(b\) with whatever way you have been taught (Quadratic Formula etc.)

Once you find \(b\), you know it is equal to \(x\). Then, simply sub in \(b\) to find \(x\)
\(x^2=b\)
\(x=\pm\sqrt{b}\)

Answer 3

That should help, @minimoo

Answer 4

\[x^4 -41x^2 =-400\]\[(x^4-41x^2)+400=0\]\[\left(x^4 -41x^2 +\frac{1681}{4}\right) +400 -\frac{1681}{4}=0\]\[\left(x^2-\frac{41}{2}\right)^2 -\frac{81}{4}=0\]\[x^2-\frac{41}{2} = \pm \sqrt{\frac{81}{4}}\]\[x^2 = \frac{41}{2} \pm \frac{9}{2}\]

Answer 5

I honestly think this is a much better approach, and it teaches you to quickly identify quadratics without using the quadratic formula.

Answer 6

So you did the completing the square method @Jhannybean :). I just said to use any way that person is comfortable with.

Answer 7

Oh true true. I just love completing squares. (pun intended) :P

Answer 8

so you would have \[x^2 = \frac{ 50}{2} \ ,\ \frac{32}{2}\]\[x^2 = 25 \ , \ 16 \]\[x = \pm \sqrt{25} \ ,\ \pm \sqrt{16}\]\[x = \pm 5 \ , \ \pm 4\]