find the zeros of the quadratic polynomial (8x2-4) and verify the relation between the zeros and its coefficients
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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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