Question

if a and b are the zeroes of the polynomial p(x)=x2-px+q then the value of 1/a+1/b is..
(a) q/p (b) p/q (c) p2-1 (D) q2-p/q.​

Answer 1

ANSWER:

Given:

• a and b are zeroes of p(x)
• p(x) = x^2 - px + q

To Find:

• Value of 1/a + 1/b

Solution:

We are given that,

$\implies\sf p(x)=x^2-px+q$

As, a and b are zeroes of p(x),

• Sum of zeroes = a + b = -(coefficient of x)/(coefficient of x^2) = -(-p)/1 = p -----(1)
• Product of zeroes = ab = (constant)/(coefficient of x^2) = q/1 = q -----(2)

We need to find the value of,

$\implies\sf \dfrac{1}{a}+\dfrac{1}{b}$

$\implies\sf \dfrac{a+b}{ab}$

From (1) & (2),

$\implies\sf \dfrac{a+b}{ab}$

$\implies\sf \dfrac{p}{q}$

So,

$\implies\bf \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{q}$

Hence, option b) p/q is correct.

Formula Used:

• Sum of zeroes = -(coefficient of x)/(coefficient of x^2)
• Product of zeroes = (constant)/(coefficient of x^2)