Question

find the zeros of the following quadratic polynomials and varify there relationship between the zeros and the coefficients 8x^2-4​

Answer 1

Answer:

We have: f(x) = 8x2 ˗ 4 It can be written as

8x2 + 0x ˗ 4 = 4

{ (√2x)2 ˗ (1)2 } = 4 (√2x + 1) (√2x ˗ 1)

∴ 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 1/√2 Here the coefficient of x is 0 and

the coefficient of x2 is √2...

$\fbox\red{or other method}$

Let f(x) = 8x2 – 4

Put f(x) = 0

8x2 – 4 = 0

(2√2x – 2) (2√2x + 2) = 0

Answer 2

Answer:

Find the zeros of the following