Question

if px^2 +Qx+R is a polynomial having zeroes at x=1 and -2.Find P+Q+R​

Answer 1

Answer:

Let α,β and ψ be the zeros of the polynomial f(x)=x  

2

−3px  

2

+qx−r

f(x)=x  

2

−3px  

2

+qx−r

=(x−α)(x−β)(x−ψ)

=x  

2

−(α+β+ψ)x  

2

+(αβ+βψ+ψα)x−αβψ

Equating the coefficients of x  

2

 we have

−(α+β+ψ)=−3p

α+β+ψ=3p ……….(1)

Now, it is given that α,β & ψ are in A.P. Let S be the common difference of terms of the AP.

⇒β−α=δ

⇒α=β−δ ………….(2)

ψ−β=δ

⇒ψ=β+δ ……………..(3)

Put the values of α and ψ from equation (2) and (3) in (1)

β−δ+β+β+δ=3p

⇒3β=3p

⇒β=p

⇒ p is a root of the polynomial f(x)

put x=p in f(x)

⇒0=f(p)=(p)  

3

−3p(p)  

2

+q(p)−r

p  

3

−3p  

3

+qp−r=0

−2p  

2

+qp−r=0

⇒2p  

3

=qp−r.

Step-by-step explanation: