AP EAMCET202123 Aug 2021Evening ShiftMathematicsQuadratic EquationActual
If one of the roots of the equation x^2+p x+q=0 is equal to the square of the other then
Options
- Ap (q^2-3 p )=q(p-1)
- Bp (3 p-q^2 )=p(p+1)
- Cp (3 q-p^2 )=q(q-1)
- Dp (3 q-p^2 )=q(q+1)
Correct answer
D. p (3 q-p^2 )=q(q+1)
Step-by-step solution
Given equation x^2+p x+q=0 ...(i) On comparing with a x^2+b x+c=0a=1, b=p, c=q Let , are the roots of Eq. (i) and given that = ^2 + = -b a =-p and = c a =q ^2+ =-p and ^2 =q and ^3=q ...(ii) and =(q)^ 1 / 3 On taking cube both sides ( ^2+ )^3=(-p)^3 ^6+ ^3+3 ^3 ( ^2+ )=-p^3 q^2+q+3 q(-p)=-p^3 [using Eq. (ii) ]q^2+q-3 p q=-p^3 -p^3+3 p q=q^2+q p (3 q-p^2 )=q(q+1)