# Solving a Polynomial Equation using quadratic techniques.

1. Mar 21, 2006

### alancj

Ok, I’m having trouble solving this equation:

-x^4+200=102x^2

The solutions I got for it are x=10 and x=-10 which both work as solutions but the last one I got, x= the square root of 2, doesn’t work.

My textbook offers the only advice for doing this type of problem by using “Quadratic Techniques.” So I went through the examples and tried to solve it. Here’s what I did:

-x^4+200=102x^2
Original equation

(-x^2)^2-102(x^2)+200=0
Made it equal to zero and then wrote it in quadratic form.

(x^2-100)(x^2-2)=0
I factored the trinomial.

(x-10)(x+10)(x^2-2)
Factored the difference of squares. I left (x^2-2) as is.

So then I set each of the above there factors to zero and soled for x. And I got for my answers:
X=10
X=-10
X=square root of 2

What am I doing wrong here?

Thanks,
Alan

2. Mar 21, 2006

### HallsofIvy

Staff Emeritus
No, 10 and -10 are not solutions to -x^4+200=102x^2 either. If x= 10, the equation becomes -10000+ 200= 10200 which is not true. The left side is -9800, not 10200. You dropped the "-".
-x4 is not equal to (-x2)2), it is equal to
-(x2)2.

I would recommend multiplying -x4- 102x2+ 200= 0 by -1 to make it x4+ 102x2- 200= 0. However, the equation u2+ 102u- 200= 0 does not have integer or rational solutions.

3. Mar 21, 2006

### alancj

So... what does that mean? I'm going to have imaginary or irrational? Joy. Now that I'm trying to do it again after multiplying by -1 I can't seem to factor it. Argh. Any hints as to how I'm supposed to do this?
Thanks,
Alan

4. Mar 21, 2006

### d_leet

5. Mar 21, 2006

### alancj

I just tried the quadratic formula and I got -51 plus or minus the square root of 2801. I checked it against my original formula and it doesn't work.
Basically here's what I did.
-x^4+200=102x^2
Original
(x^2)^2+102(x^2)-200=0
Moved things around, put in quadratic form and multiplied by -1.
I then defined a=1 b=102 and c=-200.
Then I went through the formula and got what I stated above. Maybe I defined the variables wrong?

I also tried completing the square and that didn't seem to work either. At the end I had to find the square root of +/- "imaginary number" -2. The imaginary was already there from a previous step. So finding the root of -2 would give me another "i" and that would give me "i" times "i" which is -1.
So x=+/- -1 square root of 2 was the answer that didn't work.

I'm probably like 1 obvious mistake away from getting these right. :grumpy:

-Alan

6. Mar 21, 2006

### d_leet

How does that not work? That should be right. And remember also that that is what x2 is equal to not what x is equal to...

7. Mar 22, 2006

### alancj

Damn. I thought it was = to x. Ok, so to find the value of x I would have to find the square root of both sides, and that would then be:

x=i sqrt(51) +/- 4rth-rt(2801)

every time I try calculating that it is never the same as finding the square root of the x^2= one with the calculator. I'm going insane!
I get in decimal (+/-):
x^2= ~1.924 or ~ -103.924
sqrt of x^2= ~1.387 or ~ 10.194i this works for the value of x!

But to find the exact answer I want to say that x=i sqrt(51) +/- 4rth-rt(2801) but when I calculate this I get ~0.13349 or ~14.4163

I'm so confused. Do I just leave it at x^2? The problem just says "Solve:..." Should I just give decimal answers for x? Or is there a way to find the exact x= answer?

Thanks,
Alan

8. Mar 22, 2006

### Hurkyl

Staff Emeritus
$$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$$

9. Mar 22, 2006

### Integral

Staff Emeritus
My sugestion would be to let $t = x^2 [/tex] then express your original equation in terms of t. You now have a easy to deal with quadratic. Once you find the roots of the quadratic in terms of t, take the square root to get x. 10. Mar 24, 2006 ### alancj Then what does equal $$\sqrt{a + \sqrt {b}}$$ ? I would think that there would be a way to find the exact answer to this problem. I mean $$\sqrt {\sqrt {b}}$$ should be $$\sqrt[4]{b}$$ but if I can't just add $$\sqrt {a}$$ to that then I'm lost. Unless this kind of problem is just usually left in the $${x^2=}$$ form? Last edited: Mar 24, 2006 11. Mar 24, 2006 ### ksinclair13 Well, let a = 4 and b = 16. According to your logic, the answer would be 4, but the answer is actually [itex]\sqrt{8}$.

Edit: I'm sorry about my wording. I was just showing that $\sqrt{a + \sqrt {b}}$ isn't equal to $\sqrt[4]{b} + \sqrt{a}$, but after re-reading your post, it seems that you know that.

Last edited: Mar 24, 2006
12. Mar 24, 2006

### alancj

What are you talking about? According to what logic?

13. Mar 24, 2006

### alancj

To clarify what I'm asking...

If $$x^2= -51 \pm\sqrt{2801}$$

Then $$x= \sqrt {-51 \pm\sqrt{2801}}$$

Ok, can the last one can be simplified? Or should it be left like that? If it can be simplified how would one go about doing so? What rules apply?

Thanks,
Alan

Last edited: Mar 24, 2006
14. Mar 25, 2006

### HallsofIvy

Staff Emeritus
Since 2801 is "square free" there isn't any way to simplify that. Leave it as $$x= \pm \sqrt {-51 \pm\sqrt{2801}}$$.
Those are the four roots of the original equation.

15. Mar 25, 2006

### arildno

Note that two of your roots are imaginary, and two are real.

You didn't state if you were after only the real roots of your equations; if that is the case, then you've got two solutions.

16. Mar 27, 2006

### benorin

Numbers of of the form $$\frac{a+\sqrt{b}}{c}$$ are called quadratic surds (or quadratic irrationals) for squarefree b, a,b, and c are integers. I think I recall reading a theorem out of A Course of Pure Mathematics by GH Hardy which stated that the square-root of a quadratic surd is a quadratic surd... but hey, the link I have to that text online won't load proper.

17. Mar 27, 2006

### alancj

The test question was simply "Solve: <problem>" I only got two solutions in the first place, and I really don't know how you would get more than two. I got one real and one imaginary.

I decided to give the approximate decimal answers in the above posts as well as the exact answer:
$$x= \sqrt {-51 \pm\sqrt{2801}}$$

Anyway...

I hope I get all this right because I have spent way too much time on it to get it wrong! I also hope I didn't flop on the other 29 questions which were largely pretty easy but there is always a chance for stupid mistakes.

I sent it in by snail-mail today (it's a correspondence course) and the last Unit Exam that I sent in took all of six weeks for those bums to grade. The regular exams -chapter exams- are multiple choice and they can be submitted and graded online which is nice.

I'm trying to get the best possible grades on these exams because the first third of this class I did pretty quick and I made a lot of needless mistakes. If I want an A in this class I half to get very high scores to make up for the not so stellar ones I got before. Also any future career paths would likely be in an engineering field which would necessitate quite a bit of math, for which this stuff would be a foundation for more advanced curriculum.

Thanks for all the help guys,
Alan

18. Mar 27, 2006

### arildno

Read HallsofIvy's post again to find your two missing roots.

19. Mar 27, 2006

### alancj

Crap... I didn't see the $$\pm$$ in front of the equation. I hope the person grading doesn't notice that it's missing. In that case then I can see how there would be 4 solutions. Oh, well I already sent it in... Damn

Last edited: Mar 27, 2006
20. Mar 30, 2006

### Plastic Photon

10 and -10 are correct since -x^4=x^4, unless of course it is meant to mean-(x^4).

It is possible to graph each side of the equation and find the intersection.