If αand β are two real roots of the equation x3−px2+qx+r=0 satisfying the relation αβ+1=0, then find the value of r2+pr+q.
Solution: For x3+px2+qx+r=0
Sum of the roots α+β+γ=−p . . . (1)
αβ+βγ+γα=q . . . (2)
αβγ=−r . . . (3) (Given αβ=−1)
From equation 3,
(−1)γ=−r
⇒γ=r
Substituting γ in equation 2
−1+(α+β)r=q
⇒α+β=q+1r . . . (4)
Substitute α+β+γ=−p
⇒q+1r+r=−p
⇒q+1+r2=−pr
⇒r2+pr+q+1=0
⇒r2+pr+q=−1