Prove that if a quadratic has real coefficients such that the sum of the coefficients is zero, then the quadratic has real roots.
\(\text{If $p(x)$ isn't monic, then normalize it so it is. $\\$This won't affect the roots or the sum of the coefficients being 0}\\ p(x) = x^2 + b x -(1+b)\\ p(x) = (x-1)(x+1+b) \text{ with roots}\\ x=1,~x=-(1+b) \text{, both of which are real}\)
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