Question

Find the zeroes of the quadratic polynomial (8x² – 4) and verify the relation between the zeroes and the coefficients.

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

8x² – 4

Answer 1

We have: 

f(x) = 8x² – 4 

It can be written as 8x² + 0x – 4 

= `4{(sqrt(2)x)^2 - (1)^2}` 

= `4(sqrt(2)x + 1) (sqrt(2)x - 1)` 

∴ f(x) = 0 ⇒ `(sqrt(2)x + 1) (sqrt(2)x - 1) = 0` 

⇒ `(sqrt(2)x + 1) = 0` or `sqrt(2)x - 1 = 0` 

⇒ `x = (-1)/sqrt(2)` or `x = 1/sqrt(2)` 

So, the zeroes of f(x) are `(-1)/sqrt(2)` and `1/sqrt(2)` 

Here the coefficient of x is 0 and the coefficient of x² is `sqrt(2)` 

Sum of zeroes = `(-1)/sqrt(2) + 1/sqrt(2)`

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

= `0/sqrt(2)`

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

Product of zeroes = `(-1)/sqrt(2) xx 1/sqrt(2)`

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

= `(-4)/8`

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