SOLUTION: if the roots of the quadratic equation {{{px^2+qx+r=0}}} differ by 4, show that {{{q^2=4p(4p+r)}}}. ThNK yOU for your help

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: if the roots of the quadratic equation {{{px^2+qx+r=0}}} differ by 4, show that {{{q^2=4p(4p+r)}}}. ThNK yOU for your help      Log On


   

Question 134785: if the roots of the quadratic equation px%5E2%2Bqx%2Br=0 differ by 4, show that q%5E2=4p%284p%2Br%29. ThNK yOU for your help
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Using the quadratic formula, the roots of px%5E2%2Bqx%2Br=0 are:

x+=+%28-q+%2B-+sqrt%28+q%5E2-4pr+%29%29%2F%282p%29+ and these two roots differ by 4, so:



2%28sqrt%28+q%5E2-4pr+%29%29%2F%282p%29=4

sqrt%28+q%5E2-4pr+%29%2Fp=4

sqrt%28+q%5E2-4pr+%29=4p

q%5E2-4pr=16p%5E2

q%5E2=16p%5E2%2B4pr

q%5E2=4p%284p%2Br%29

In general, if the roots differ by d, then q%5E2=dp%28dp%2Br%29