Express q as a function of p, given that one root of x^2 + px + q = 0 is twice the other.
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x2 = 2*x1 (x-x1)(x-2*x1) = x^2+p*x+q x^2-(2*x1+x1)*x+2*(x1^2) = x^2+p*x+q x^2-(3*x1)*x+2*(x1^2) = x^2+p*x+q So: p = -3*x1 q = 2*(x1^2) or: x1 = (-1/3)p so: q = (2/9)*(p^2)
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