Question

If a and b are the zeros of the polynomial p(x) = x^2 - px +q, then find the value of 1/a + 1/b​

Answer 1

ANSWER:

Given:

• f(x) = x² - px + q
• Zeroes of f(x) = a & b

To Find:

• Value of 1/a + 1/b

Solution:

We are given a polynomial f(x) = x² - px + q.

The zeroes of f(x) are 'a' and 'b'.

So,

⇒ Sum of zeroes = -(Coefficient of x)/(Coefficient of x²)

⇒ Sum of zeroes = -(-p)/1 = p --------(1)

And,

⇒ Product of zeroes = (Constant)/(Coefficient of x²)

⇒ Product of zeroes = (q)/1 = q --------(2)

Now,

⇒ 1/a + 1/b

Taking LCM,

⇒ 1/a + 1/b = (b + a)/ab

⇒ 1/a + 1/b = (a + b)/ab

Here,

⇒ (a + b) = Sum of zeroes

And,

⇒ ab = Product of zeroes

So, from (1) & (2),

⇒ 1/a + 1/b = (p)/(q)

⇒ 1/a + 1/b = p/q

So, the value of 1/a + 1/b is p/q.

Formula Used:

• Sum of zeroes = -(Coefficient of x)/(Coefficient of x²)
• Product of zeroes = (Constant)/(Coefficient of x²)

Answer 2

Step-by-step explanation:

see the picure. okay?

answer in it