Question

If zeros of the polynomial $f(x)=x^{3}-3px^{2}+qx-r$ are in A.P., then
(a) $2p^{3}=pq-r$
(b) $2p^{3}=pq+r$
(c) $p^{3}=pq-r$
(d) None of these

Answer 1

Answer:

Step-by-step explanation:

SOLUTION :  

The correct option is c : qp - r = 2p³ .

Let α,β,γ are the three Zeroes of the  polynomial.

Given : f(x) = x³ - 3px² + qx - r

α = a - d , β = a , γ = d

On comparing with ax³ + bx² + cx + d ,

a = 1 , b = -3p , c = q , d = -r

Sum of zeroes of cubic POLYNOMIAL = −coefficient of x² / coefficient of x³

α + β + γ = −b/a

(a - d) + a + (a + d) = -(-3p)/1

a + a + a = 3p

3a = 3p

a =(3/3)p

a = p

Since, a is a zero of the polynomial f(x) , Therefore, f(a) = 0

a³ - 3pa² + qa - r = 0

On substituting a = p ,

p³ - 3p(p)² + qp - r = 0

p³ - 3p³ + qp - r = 0

-2p³ + qp - r = 0

qp - r = 2p³

Hence, the correct option is (c) : qp - r = 2p³ .

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Answer 2

Answer:

plz refer to the given attachment...

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