Question 1209115
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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<ul><li>The x coefficients on the left and right sides are -(p+q) and b respectively. Therefore b = -(p+q).</li><li>The constant terms on the left and right sides are p*q and c respectively. Therefore c = p*q.</li></ul>Side note: This is the quadratic case of <a href="https://mathworld.wolfram.com/VietasFormulas.html">Vieta's Formulas</a>.


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.
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