Question

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.

6x² – 3 – 7x

Find the zeroes of the quadratic polynomial 6x² – 3 – 7x and verify the relationship between the zeroes and the coefficients.

Answer 1

6x² – 3 – 7x

= 6x² – 7x – 3

= 6x² – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3)(3x + 1)

= `2(x - 3/2)3(x+1/3)`

For p(x) = 0 we have,

Either (3x + 1) = 0

`x = -1/3`

or (2x – 3) = 0

`x = 3/2`

Thus, the zeroes of 

6x² – 3 – 7x are `-1/3  "and"  3/2` 

⇒ Sum of the zeroes = `"-Coefficient of x"/("Coefficient of" x^2)`

⇒ `-1/3 + 3/2= (- (-7))/6`

⇒ `7/6 = 7/6`

Product of the zeroes = `"Constant term"/("Coefficient of "x^2)`

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

⇒ `-1/2 = -1/2`

Thus, the relationship between the zeroes and coefficients in the polynomial 6x² – 3 – 7x is verified.

Answer 2

Given, quadratic polynomial:

p(x) = 6x² – 3 – 7x

For zeroes of polynomial, put p(x) = 0

∴ 6x² – 7x – 3 = 0

By splitting the middle term,

6x² – 9x + 2x – 3 = 0

3x(2x – 3) + 1(2x – 3) = 0

(2x – 3) (3x + 1) = 0

∴ 2x – 3 = 0 and 3x + 1 = 0

or `x = 3/2` and `x = - 1/3`

Therefore, `α = 3/2` and `β = - 1/3` are the zeroes of the given polynomial.

Verification:

Sum of zeroes = α + β

= `3/2 + (-1/3)`

= `3/2 - 1/3`

= `(9 - 2)/6`

= `7/6`

= `-(("Coefficient of"  x))/("Coefficient of"  x^2)`

and Product of zeroes = αβ

= `(3/2)(-1/3)`

= `-1/2`

= `"Constant term"/("Coefficient of"  x^2)`

Hence Verified.