Question

verify relationship of co efficient and zero of 8s square -4s​

Answer 1

Answer:

Solution:

Recall the identity

a²-b² = (a+b)(a-b)

Using it , we can write :

8x²-4

= 4(2x²-1)

= 4[(√2x)²-1²]

= 4(√2x+1)(√2x-1)

So, the value of 8x²-4 is zero

when x = -1/√2 and 1/√2

verification:

Compare 8x²-4 with ax²+bx+c ,

we get

a = 8, b = 0 , c = -4

i ) Sum of the zeroes

= -1/√2+1/√2

= 0

= -(coefficient of x)/(coefficient of x²)

ii) Product of the zeroes = (-1/√2)×(1/√2)

= -1/2

= (constant term)/(coefficient of x²)

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