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Question
Find the zeroes of the quadratic polynomial `(8x^2 ˗ 4)` and verify the relation between the zeroes and the coefficients
Solution
We have:
`f(x)=8x^2-4`
It can be written as `8x^2+o x -4`
=`4{(sqrt2x)^2-(1)^2}`
=`4(sqrt2x+1) (sqrt2x-1)`
∴ `f(x)=0⇒ (sqrt2x+1) (sqrt2x-1)=0`
⇒ `(sqrt2x+1)=0 or sqrt2x-1=0`
⇒ `x=(-1)/sqrt2 or x=1/sqrt2`
So, the zeroes of f(x) are `(-1)/sqrt2 and 1/sqrt2`
Here the coefficient of x is 0 and the coefficient of `x^2` is `sqrt2`
Sum of zeroes = `-1/sqrt2+1/sqrt2=(-1+1)/sqrt2=0/sqrt2=-(("Coefficent of x"))/(("Coefficient of" x^2))`
Product of zeroes=`-1/sqrt2xx1/sqrt2=(-1xx4)/(2xx4)=-4/8=("Constant term")/(("Coefficient of" x^2))`
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