+0  
 
-1
110
2
avatar+127 

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

 Jun 28, 2020
 #1
avatar+781 
+1

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
avatar+8342 
0

(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

49 Online Users