SOLUTION: If the roots of the quadratic equation {{{2x^2 + x +6 = 0}}} are {{{alpha}}} and {{{beta}}}, find the quadratic equation whose roots are {{{alpha + 1/(2*beta)}}} and {{{beta + 1/(2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: If the roots of the quadratic equation {{{2x^2 + x +6 = 0}}} are {{{alpha}}} and {{{beta}}}, find the quadratic equation whose roots are {{{alpha + 1/(2*beta)}}} and {{{beta + 1/(2      Log On


   

Question 1041375: If the roots of the quadratic equation 2x%5E2+%2B+x+%2B6+=+0 are alpha and beta, find the quadratic equation whose roots are alpha+%2B+1%2F%282%2Abeta%29 and beta+%2B+1%2F%282%2Aalpha%29.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2+%2B+x+%2B6+=+0 <===> x%5E2+%2B+x%2F2+%2B3+=+0
Hence alpha%2Bbeta+=+-1%2F2 and alpha%2Abeta+=+3
For the new quadratic equation:

The coefficient of x = -(Sum of the roots)
= -%28alpha+%2B+1%2F%282%2Abeta%29+%2B+beta+%2B+1%2F%282%2Aalpha%29%29.
=-
= -%28+-1%2F2+%2B%281%2F2%29%2A%28%28-1%2F2%29%2F3%29%29 = 7/12
The constant in the quadratic equation = product of roots
= %28alpha+%2B+1%2F%282%2Abeta%29%29%28beta+%2B+1%2F%282%2Aalpha%29%29%29.
=alpha%2Abeta+%2B1+%2B+1%2F%284%2Aalpha%2Abeta%29+=+3%2B1%2B1%2F12+=+49%2F12.
===> the quadratic equation is x%5E2%2B+%287%2F12%29x+%2B+49%2F12+=+0, or after clearing fractions,
12x%5E2+%2B+7x+%2B+49+=+0.