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 Check out our web site! http://mathforum.org/dr.math/
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