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Question
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients:
x2 – 2x – 8
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Solution
By factorization method:
x2 - 2x - 8
⇒ x2 - 4x + 2x - 8 = 0
⇒ x(x - 4) + 2(x - 4) = 0
⇒ x(x - 4) + 2(x - 4) = 0
⇒ (x - 4) (x + 2) = 0
⇒ x - 4 = 0, x + 2 = 0
⇒ x = 4, x = -2
For p(x) = 0, we must have (x - 4) (x + 2) = 0 Either x - 4 = 0
x = 4
or x + 2 = 0
x = -2
∴ The zeroes of x2 - 2x - 8 are 4 and -2
Now,
= Sum of the zeroes `="-Coefficient of x"/"Coefficient of x"`
`-2+4=(-(-2))/1`
2 = 2 (L.H.S = R.H.S)
Product of the zeroes `="Constant term"/("Coefficient of "x^2)`
`-2xx4=(-8)/1`
`-8 = -8` (L.H.S = R.H.S)
Thus, the relationship between the zeroes and the coefficients in the polynomial x2 – 2x – 8 is verified.
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