Question

Verify the relationship between the zeroes and the coefficients of the quadratic polynomial 9x² – 25.

Answer 1

Given polynomial = 9x² – 25

To find zeroes

9x² – 25 = 0

= (3x)² – (5)² = 0

= (3x – 5)(3x + 5) = 0   ...[∵ a² – b² = (a – b)(a + b)]

3x – 5 = 0

⇒ 3x = 5

⇒ `x = 5/3`

3x + 5 = 0

⇒ 3x = –5

⇒ `x = -5/3`

∴ `α = 5/3, β = (-5)/3`

Identify co-efficients:

9x² – 25   ...[On comparison]

= ax² + bx + c

a = 9, b = 0, c = –25

Verify relationship:

`α + β = (-b)/a`

LHS = α + β

= `5/3 - 5/3`

= 0

RHS = `(-b)/a`

= `0/9`

RHS = 20

`αβ = c/a`

LHS = αβ

= `(5/3)((-5)/3)`

LHS = RHS

LHS = `(5/3)((-5)/3)`

= `(-25)/9`

RHS = `c/a = (-25)/9`

∴ LHS = RHS

Hence, verified both relations.