Question

Find the zeroes of the polynomial p(x) = 6x² + 13x – 5 and verify the relationship between its zeroes and the coefficients.

Answer 1

p(x) = 6x² + 13x – 5

The zeros of the polynomial are the values of x, where p(x) = 0.

Now, putting values of p(x) = 0

6x² + 13x – 5 = 0

⇒ 6x² + 15x – 2x – 5 = 0

⇒ 3x(2x + 5) – 1(2x + 5) = 0

⇒ (2x + 5) (3x – 1) = 0

If 2x + 5 = 0, then x = `(-5)/2`

or 3x – 1 = 0, then x = `1/3`

Relation between its zeroes and the coefficients.

Sum of zeroes = α + β = `(-"(Coefficient of x)")/"Coefficient of x"^2`

= `((-5)/2) + (1/3) = (-13)/6`

= `((-5 xx 3) + (1 xx 2))/6 = (-13)/6`

= `(-15 + 2)/6 = (-13)/6`

= `(-13)/6 = (-13)/6`

Product of zeroes = αβ = `"Constant term"/"Coefficient of x"^2`

= `((-5)/2) xx (1/3) = (-5)/6`

`= (-5)/6 = (-5)/6`