## Descartes and Real Solutions

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Szechuan
Posts: 6
Joined: Thu Aug 04, 2011 7:35 pm
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### Descartes and Real Solutions

Hello,

I am struggling with the following question and was hoping for some assistance:

How many real solutions does this equation have?: (x^4) - (5x^2) +6 = 0
I inserted the brackets to make it easier to read. The solution in the book reads '4', with no other details whatsoever.

The first step I did was to determine positive and negative case and look for sign changes:

Pos: (x^4) - (5x^2) + 6 = 0
Two changes:
One from x^4 to 5x^2
And one from 5x^2 to 6

Neg: (-x^4) - (-5x^2) + 6 = 0
Which becomes (-x^4) + 5x^2 + 6 = 0
One change: -x^4 to +5x^2.

So the total should be 3 correct? I have tried doing this several times and can't figure out what I am missing.

In addition, the Purplemath page on the Rule of Signs has some quirks that are confusing me, copied below:

Code: Select all

`Now I look at f(–x) (that is, having changed the sign on x, so this is the "negative" case):      f (–x) = (–x)^5 – (–x)^4 + 3(–x)^3 + 9(–x)^2 – (–x) + 5               = –x^5 – x^4 – 3x^3 + 9x^2 + x + 5`

Shouldn't the negative case read = -x^5 plus x^4 - 3x^3 minus 9x^2 + x + 5?

Actually, I just had a eureka moment: Is it the -X being squared/cubed that is affecting the signs?
For example:
= -x^5 - (-x)^4
= -x^5 - (-x * -x * -x * -x)
= -x^5 - (x^2 * x^2)
= -x^5 - (x^4)
And thus there is not actually a double-negative sign and the term remains a subtraction?

Szechuan
Posts: 6
Joined: Thu Aug 04, 2011 7:35 pm
Contact:

### Re: Descartes and Real Solutions

Haha, I just redid the problem and had no trouble once I paid attention to the exponents affecting -x

Guess this one's solved?

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
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Szechuan wrote:Haha, I just redid the problem and had no trouble once I paid attention to the exponents affecting -x

Guess this one's solved?

If you got "four solutions: two positive and two negative", then yes, this is solved.