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Question
Find the zeroes of the quadratic polynomial `f(x) = x^2 + 3x ˗ 10` and verify the relation between its zeroes and coefficients.
Solution
We have:
`f(x) = x^2 + 3x ˗ 10`
=` x^2 + 5x ˗ 2x ˗ 10`
= x(x + 5) ˗ 2(x + 5)
= (x ˗ 2) (x + 5)
∴ f(x) = 0 ⇒ (x ˗ 2) (x + 5) = 0
⇒ x ˗ 2 = 0 or x + 5 = 0
⇒ x = 2 or x = −5.
So, the zeroes of f(x) are 2 and −5.
Sum of zeroes= `2+(-5)=-3=(-3)/1="(-coefficient of x)" /(("Coefficent of "x^2))`
Product of zeroes =`2xx(-5)=-10=(-10)/1= "Constant term"/(("Coefficcient of "x^2 ))`
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