Question

Find the zeros of the quadratic polynomials 8x² - 4 and verify the relationship between the zeros and the coefficients.

Answer 1

Let f(x) = 8x² – 4

= 4 ((√2x)² – (1)²)

= 4(√2x + 1)(√2x – 1)

To find the zeroes, set f(x) = 0

(√2x + 1)(√2x – 1) = 0

(√2x + 1) = 0 or (√2x – 1) = 0

x = (-1)/√2 or x = 1/√2

So, the zeroes of f(x) are (-1)/√2 and x = 1/√2

Again,

Sum of zeroes = -1/√2 + 1/√2 = (-1+1)/√2 = 0

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = -1/√2 x 1/√2 = -1/2 = -4/8

= c/a

= Constant term / Coefficient of x²