Question

find the find the zeros of the polynomial quadratic polynomial and verify the relationship between the zeros and coefficients.
8x²-4​

Answer 1

Answer:

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

Step-by-step explanation:

hope it will be helpful to you dear friend

Answer 2

Answer:

Let f(x)=8x2−4

= 4((2x)2−(1)2)

=4(2x+1)(2x−1)

to find the zeroes, Let f(x)=0

(2x+1)(2x−1)=0

(√2x+1)=0 or (√2x−1)=0

x=2(−1) or x=21

So the zeroes of f(x) are  2−1 and x=21

Again 

Sum of zeroes = 2−1+21=2