Question

find the zeros of 8x^2 - 4 and verify​

Answer 1

Answer:

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)

HOPE THIS HELPS YOU!!!

Answer 2

8x²- 4 = 4(2x² - 1) = 4[ (x√2)² - (1)² ] = 4(x√2 + 1)(x√2 - 1)

Therefore,

(x√2 + 1) = 0                           and                           (x√2 - 1) = 0

x = -1/√2                                  and                            x = 1/√2

Hence the zeroes are

α = -1/√2

β = 1/√2

α+β = -b/a

-1/√2 + 1/√2 = 0/8

0 = 0

αβ = c/a

-1/√2*1/√2 = -4/8

-1/2 = -1/2

Hope this is helpful to you

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