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