Question

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

x² – 2x – 8

Answer 1

By factorization method:

x² - 2x - 8

⇒ x² - 4x + 2x - 8 = 0

⇒ x(x - 4) + 2(x - 4) = 0

⇒ x(x - 4) + 2(x - 4) = 0

⇒ (x - 4) (x + 2) = 0

⇒ x - 4 = 0, x + 2 = 0

⇒ x = 4, x = -2

For p(x) = 0, we must have (x - 4) (x + 2) = 0 Either x - 4 = 0

x = 4

or x + 2 = 0

x = -2

∴ The zeroes of x² - 2x - 8 are 4 and -2

Now,

= Sum of the zeroes `="-Coefficient of x"/"Coefficient of x"`

`-2+4=(-(-2))/1`

2 = 2 (L.H.S = R.H.S)

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

`-2xx4=(-8)/1`

`-8 = -8` (L.H.S = R.H.S)

Thus, the relationship between the zeroes and the coefficients in the polynomial x² – 2x – 8 is verified.