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प्रश्न
Find the zeroes of the quadratic polynomial (8x2 – 4) and verify the relation between the zeroes and the coefficients.
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
8x2 – 4
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उत्तर
We have:
f(x) = 8x2 – 4
It can be written as 8x2 + 0x – 4
= `4{(sqrt(2)x)^2 - (1)^2}`
= `4(sqrt(2)x + 1) (sqrt(2)x - 1)`
∴ f(x) = 0 ⇒ `(sqrt(2)x + 1) (sqrt(2)x - 1) = 0`
⇒ `(sqrt(2)x + 1) = 0` or `sqrt(2)x - 1 = 0`
⇒ `x = (-1)/sqrt(2)` or `x = 1/sqrt(2)`
So, the zeroes of f(x) are `(-1)/sqrt(2)` and `1/sqrt(2)`
Here the coefficient of x is 0 and the coefficient of x2 is `sqrt(2)`
Sum of zeroes = `(-1)/sqrt(2) + 1/sqrt(2)`
= `(-1 + 1)/sqrt(2)`
= `0/sqrt(2)`
= `(-("Coefficent of" x))/(("Coefficient of" x^2))`
Product of zeroes = `(-1)/sqrt(2) xx 1/sqrt(2)`
= `(-1 xx 4)/(2 xx 4)`
= `(-4)/8`
= `("Constant term")/(("Coefficient of" x^2))`
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