# Polynomial with 3 unknown variables

1. Aug 31, 2009

### Paulo Serrano

1. The problem statement, all variables and given/known data

(4m + 3n)x2 – 5nx + (m – 2) = 0

The sum of the roots is 5/8
The product of the roots is 3/32

What is M + N?

2. Relevant equations

3. The attempt at a solution

This is the actual solution. I just don't understand it.

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Why is it that although it says the SUM is equal to 5/8 the solution above has has things being divided? How do I find the roots of this equation?

2. Aug 31, 2009

### rock.freak667

For ax2+bx+c=0 (or x2+(b/a)x+c/a=0) if the roots are α and β, then x2+(b/a)x+c/a≡ (x-α)(x-β)

The right side works out to be x2-(α+β)x+ (αβ). So equating coefficients we get

α+β=-b/a and αβ=c/a

or the sum of the roots= -b/a
the product of the roots=c/a

3. Aug 31, 2009

### VietDao29

If you want some further reading, you can have a look here. It's called http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas" [Broken].

Last edited by a moderator: May 4, 2017
4. Sep 1, 2009

### Paulo Serrano

Is that just a rule that I should memorize then? I couldn't really work this out using the actual numbers, could I?

5. Sep 1, 2009

### VietDao29

Yes, this formula is pretty well-known, and you should memorize it. It's not very hard to memorize, yes?

This formula comes very handy, when you need to work with the sum and/or product of the root of a polynomial, without the need to know what the roots are (since it's calculated based on the coefficients of the polynomial itself).

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For example, you are ask to evaluate the sum of the 2 roots of the following equation:

x2 + x - 2 = 0

The first way is to find the 2 roots of this equation, which turn out to be -2, and 1. And take the sum: -2 + 1 = -1.

The second way is to apply the formula:
x1 + x2 = -b/a = -1/1 = -1.

Err, I don't really get what you mean. What do you mean by 'actual numbers'?

6. Sep 1, 2009

### HallsofIvy

Staff Emeritus
If you don't like memorizing, think it out: If $\alpha_1$ and $\alpha_2$ are roots of $x^2+ bx+ c= 0$, then we must have $(x- \alpha_1)(x- \alpha_2)= x^2+ bx+ c= 0$ for all x. Multiplying that first product, $x^2- (\alpha_1+ \alpha_2)x+ \alpha_1\alpha_2)= x^2+ bx+ c$ and, since that is true for all x, taking x= 0 gives $\alpha-1\alpha_2= c$ and then, taking x= 1, $1^2- (\alpha_1+ \alpha_2)(1)+ c= 1^2+ b(1)+ c$ so $\alpha_1+ \alpha_2= -b$.

7. Sep 2, 2009

### Paulo Serrano

I meant that if I tried solve this without knowing those equations it would be near impossible...Anyway, thanks again guys. Learned a little more. :)

8. Sep 9, 2009