Question

# If $α,β,γ$ are the roots of $x_{3}+px_{2}+qx+r=0$, then $∑α_{2}(β+γ)$ is

**A**

## $3r+pq$

**B**

## $3r−pq$

**C**

## $pq−3r$

**D**

## $pq+r$

Medium

Solution

Verified by GMS

Correct option is B)

## Given: $α,β,γ$ are the roots of $x_{3}+px_{2}+qx+r=0$

We have

$α+β+γ=−p$$αβ+βγ+γα=q$$αβγ=−r$

Now,

$∑α_{2}(β+γ)=(α_{2}β+α_{2}γ)+(β_{2}γ+β_{2}α)+(γ_{2}α+γ_{2}β)$

$=(α+β+γ)(αβ+βγ+γα)−3αβγ$

$=−pq+3r$

$=3r−pq$

$α+β+γ=−p$

$αβ+βγ+γα=q$

$αβγ=−r$

Now,

$∑α_{2}(β+γ)=(α_{2}β+α_{2}γ)+(β_{2}γ+β_{2}α)+(γ_{2}α+γ_{2}β)$

Now,

$∑α_{2}(β+γ)=(α_{2}β+α_{2}γ)+(β_{2}γ+β_{2}α)+(γ_{2}α+γ_{2}β)$

$=(α+β+γ)(αβ+βγ+γα)−3αβγ$

$=−pq+3r$

$=3r−pq$