If α,β are the roots of equation a(x2−1)+2bx=0, then the equation whose roots are 2α−1β & 2β−1α is
ax2+6bx−9a=0
Given that α & β are the roots of the equation a(x2−1)+2bx=0.
⇒α+β=−2ba & αβ=−1
We need to find the equation whose roots are 2α−1β & 2β−1α.
Sum of roots:
S=2α−1β+2β−1α
S=2(α+β)−(1α+1β)
S=2(α+β)−α+βαβ
S=−4ba+−2ba=−6ba
⇒S=−6ba
Now, Product of Roots:
P=2(α−1β)(2β−1α)
P=4αβ−2−2+1αβ
P=−9
The required quadratic equation would be of the form:
x2−Sx+P=0
x2−(−6ba)x−9=0
ax2+6bx−9a=0