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Theorem About Sum and Product of Quadratic Roots


Date: 03/07/98 at 14:55:59
From: LAURA
Subject: HONORS ALG.2/TRIG.

Using the theorem about the sum and product of roots, explain why 1/2 
and 3/4 are NOT the roots of the equation 0 = 4x^2 + 5x + 8. Write an 
equation for which 1/2 and 3/4 are the roots.

I have tried many times but I do not understand how I am to apply the 
theorem to the question.


Date: 03/07/98 at 15:22:49
From: Doctor Sam
Subject: Re: HONORS ALG.2/TRIG.

Laura,

The theorem you refer to says that if the roots of a quadratic 
polynomial ax^2 + bx + c = 0 are P and Q, then PQ = c/a and 
P + Q = -b/a.

If 3/4 and 1/2 were roots of your equation, then (3/4)(1/2) = 8/4,
which is not true. Also, 3/4 + 1/2 would have to equal -5/4 -- also 
false.

It might help you to know where these mysterious results come from. It 
will just take a little factoring. If P and Q are solutions to a 
quadratic equation, then (x - P) and (x - Q) are factors.

If you multiply these out you will get the quadratic

    (x - P)(x - Q) = x^2 - (P + Q)x + PQ = 0.

Notice that the coefficient of x is the opposite of the sum of the 
roots, and that the constant at the end, PQ, is the product of the 
roots.

Now this polynomial has a leading coefficient of 1, and your example 
did not. But you could rewrite your original equation of

   0 = 4x^2 + 5x + 8

by dividing through by 4. This won't change its solutions, but it will 
change the way it looks:

   0 = x^2 + (5/4)x + 2.

And the sum of the roots of this polynomial has to be -5/4, and the 
product 2. In order to get a polynomial whose roots are 1/2 and 3/4, 
just use them as P and Q in the factored form above and multiply out:

     (x - 1/2)(x - 3/4) =  x^2 -(5/4)x + 3/8.

If you like, you can clear fractions by multiplying through by 8:

     8x^2 - 10x + 3.

I hope that helps.

-Doctor Sam, The Math Forum
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Associated Topics:
High School Basic Algebra
High School Exponents
High School Polynomials

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