If α+β=-2 and α3+β3=-56, then the quadratic equation whose roots are α and β is

Question

If α+β=-2 and α3+β3=-56, then the quadratic equation whose roots are α and β is (A) x2+2 x-16=0 (B) x2+2 x- 15=0 (C) x2+2 x-12=0 (D) x2+2 x-8=0

Tardigrade Admin · Accepted Answer

Given that, α+β=-2 and α3+β3=-56 ⇒ (α+β)(α2+β2-α β)=-56 ⇒ α2+β2-α β=28 Also, (α+β)2=(-2)2 ⇒ α2+β2+2 α β=4 ⇒ 28+3 α β=4 ⇒ α β=-8 ∴ Required equation is x2+2 x-8=0

Q. If $α + β = - 2$ and $α^{3} + β^{3} = - 56$ , then the quadratic equation whose roots are $α$ and $β$ is

$x^{2} + 2 x - 16 = 0$

$x^{2} + 2 x - 15 = 0$

$x^{2} + 2 x - 12 = 0$

$x^{2} + 2 x - 8 = 0$

Q. If α+β=−2 and α3+β3=−56, then the quadratic equation whose roots are α and β is

x2+2x−16=0

x2+2x−15=0

x2+2x−12=0

x2+2x−8=0

Given that, α+β=−2 and α3+β3=−56 ⇒(α+β)(α2+β2−αβ)=−56 ⇒α2+β2−αβ=28 Also, (α+β)2=(−2)2 ⇒α2+β2+2αβ=4 ⇒28+3αβ=4 ⇒αβ=−8 ∴ Required equation is x2+2x−8=0

Q. If $α + β = - 2$ and $α^{3} + β^{3} = - 56$ , then the quadratic equation whose roots are $α$ and $β$ is

$x^{2} + 2 x - 16 = 0$

$x^{2} + 2 x - 15 = 0$

$x^{2} + 2 x - 12 = 0$

$x^{2} + 2 x - 8 = 0$