If the roots of x4-10 x3+37 x2-60 x+36=0 are α, α, β, β(α

Question

If the roots of x4-10 x3+37 x2-60 x+36=0 are α, α, β, β(α<β), then 2 α+3 β-2 α β= (A) 1 (B) 0 (C) -1 (D) 4

Tardigrade Admin · Accepted Answer

We have, x4-10 x3+37 x2-60 x+36=0 since, α, α, β, β are the roots of the above equation. ∴ α+α+β+β=(-(-10)/1)=10 ⇒ α+β=5 ...(i) and (α)(α)(β)(β)=(36/1) ⇒ α2 β2=36 ⇒ α β=6 ...(ii) From Eqs. (i) and (ii), we have α(5-α)=6 ⇒ α2-5 α+6=0 ⇒ α=2,3 ⇒ β=3,2 Since α<β, hence α =2, β=3 ∴ 2 α+3 β-2 α β =2 × 2+3 × 3-2 × 2 × 3 =4+9-12=1

Q. If the roots of x4−10x3+37x2−60x+36=0 are α,α,β,β(α<β), then 2α+3β−2αβ=

1

0

-1

4

Q. If the roots of
$x^{4} - 10 x^{3} + 37 x^{2} - 60 x + 36 = 0$ are
$α, α, β, β (α < β)$ , then $2 α + 3 β - 2 α β =$