SOLUTION: Wilma and Greg were trying to solve the quadratic equation x^2 + bx + c = 0. Wilma wrote down the wrong value of b (but her value of c was correct), and found the roots to be 5 a

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Question 1209115: Wilma and Greg were trying to solve the quadratic equation
x^2 + bx + c = 0.
Wilma wrote down the wrong value of b (but her value of c was correct), and found the roots to be 5 and 15. Greg wrote down the wrong value of c (but his value of b was correct), and found the roots to be -5 and -7. What are the actual roots of x^2 + bx + c = 0?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!

If p,q are the roots then (x-p) and (x-q) are factors.
We can then say (x-p)(x-q) = x^2+bx+c

Expand out the left hand side to get
(x-p)(x-q) = x^2+bx+c
x^2-qx-px+p*q = x^2+bx+c
x^2-(p+q)x+p*q = x^2+bx+c

Compare terms
The x coefficients on the left and right sides are -(p+q) and b respectively. Therefore b = -(p+q).
The constant terms on the left and right sides are p*q and c respectively. Therefore c = p*q.
Side note: This is the quadratic case of Vieta's Formulas.

Wilma has the correct value of c, so,
c = p*q = 5*15 = 75

Greg meanwhile has the correct value of b
b = -(p+q) = -(-5 - 7) = 12

Therefore b = 12 and c = 75.
The quadratic Wilma and Greg are trying to solve is x^2+12x+75 = 0.

I'll let the student take over from here.

Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website!
.

When the leading coefficient of a quadratic equation is 1, then, according to Vieta's theorem, the sum of the roots is the coefficient at x with the opposite sign, and the product of the roots is the constant term. Wilma's roots 5 and 15 produce the correct constant term 5*15 = 75. Greg's roots produce the correct coefficient at x -(-5+(-7)) = -(-5-7) = -(-12) = 12. So, the correct equation is x^2 + 12x + 75 = 0. Its roots are complex numbers = = = .

Solved.