What are the real and complex solutions of the polynomial equation?

x^4-41x^2=-400

Question

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

mathstudent55 · Answer

Do a substitution.

Let $u=x^2$.
Then you get $u^2=x^4$.

Now substitute $u^2$ for $x^4$ and $u$ for $x^2$ in the given equation.
You will now have a quadratic equation in $u$.
Solve for $u$.

Can you try the above and show what you get?

deercult · Answer

$$u^{2}-41u=-400$$

Hmm... 

$$u^{2}-41u+400=0$$

Just making sure, am I off on the right foot?

deercult · Answer

Okay, here's what I have so far, using the quadratic formula and that equation.

$$x=\frac{ (41)\pm \sqrt{(-41)^{2}-4(1)(200)} }{ 2(1) }$$

deercult · Answer

@mathstudent55

mathstudent55 · Answer

You are very close.
Just remember, this is not x. Also, c = 400, not 200.
You are now solving the new equation for u.
Can you get values for u from your expression?

mathstudent55 · Answer

$u^{2}-41u+400=0$

$u = \dfrac{-(-41) \pm \sqrt{(-41)^2 - 4(1)(400)}}{2(1)} $

deercult · Answer

Oops, this whole time I was doing it in the wrong place and got the numbers confused, sorry!
Now I have:

$$u=\frac{ 41\pm \sqrt{1681-1600}}{ 2 }$$

1681-1600 is 81, and the square root of 81 is 9, leaving,

$$u=\frac{ 41\pm9 }{ 2 }$$

so far, yes?

mathstudent55 · Answer

Excellent.
Just the same I got.

$u = \dfrac{41 \pm \sqrt{1681 - 1600}}{2} $

$u = \dfrac{41 \pm \sqrt{81 }}{2} $

$u = \dfrac{41 \pm 9}{2} $

mathstudent55 · Answer

Now what two numbers does that simplify to?

deercult · Answer

41-9 = 32 / 2 = 16
41+9 = 50 / 2 = 25

Making 16 and 25 the solutions, I think

mathstudent55 · Answer

Correct.
I have to go. I'll be back in 10 minutes.

Remember that those numbers are what u is equal to.
Now substitute back.
Let x^2 equal each of those two numbers and solve the two equations for x.

$x^2 = 25$ or $x^2 = 16$

Each equation will give you two answers.
The solution to the original equation is all 4 answers.

deercult · Answer

$$x^{4}-41(25)+400=0$$
$$x^{4}-625=0$$
$$x^{4}=625$$
$$x=\pm5$$

Two answers down!

mathstudent55 · Answer

No.
Once you have 

$x^2 = 25$, 

you simply do this

$x = \pm \sqrt{25}$

$x = \pm 5$

Now do the same for $x^2 = 16$ to get the other two answers.

deercult · Answer

OH. Well, at the very least, I did get it right. :'>
I got +/- 4i for the other two!

mathstudent55 · Answer

Why i?
16 is a positive number just like 25.

$x^2 = 25$

$x = \pm 5$

$x^2 = 16$

$x = \pm 4$

deercult · Answer

Oh, that one I did goof up a little on, then, oops. I might've made a symbol mistake in my calculator. ^^;

You're an amazing helper, thank you for taking all this time to help me out!!

mathstudent55 · Answer

Since we have been doing parts and discussing them, it's easy to lose track of the flow of the steps of a complete solution, so here is a complete solution as you'd do in homework or in a quiz.

Here is the entire solution from beginning to end.

$x^4-41x^2=-400$

$x^4 - 41x^2 + 400 = 0$

Let $u = x^2$, then $u^2 = x^4$:

$u^2 - 41u + 400 = 0$

$u = \dfrac{-(-41) \pm \sqrt{(-41)^2 - 4(1)(400)}}{2(1)} $

$u = 25 $   or   $u = 16$

Substitute x back:

$x^2 = 25$   or   $x^2 = 16$

$x = \pm \sqrt{25} $   or   $x = \pm \sqrt{16} $

$x = 5$  or  $x = -5$  or  $x = 4 $  or  $x = -4$

mathstudent55 · Answer

You're welcome, and thanks for the kind words.

mathstudent55 · Answer

gtg, bye

deercult · Answer

Alright, be safe and have a good night!

mathstudent55 · Answer

ty, you too