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Let d and e denote the solutions of 4x^2+7x-1=0. Find (d-e)^2.

Jun 28, 2020

#1
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Using the quadratic equation, the two solutions to 4x^2+7x-1=0 are $$\frac{-7+\sqrt{65}}{8}\ \text{and} \frac{-7-\sqrt{65}}{8}$$. I think you can solve the rest from there.

Jun 28, 2020
#2
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(d - e)^2 = d^2 - 2de + e^2 = (d + e)^2 - 4de

By Vieta's formula, if p and q are the roots of the quadratic equation ax^2 + bx + c = 0, then p + q = -b/a, pq = c/a.

So, in this case $$\begin{cases}d + e = -\dfrac{7}4\\ de = \dfrac{-1}{4}\end{cases}$$.

You can do the rest from here.

Jun 28, 2020