Obtain a quadratic polynomial with the following conditions:
The sum of zeros = 1/3; the product of zeros = 1/2
Let α and β be the zeros of the polynomial p(x) = ax2 + bx + c.
Given, α + β =
; αβ = ![]()
We know that α + β = –
=
. So, –
=
= k, say
Thus, b = –k, a = 3k
And αβ =
=
. So, c =
= ![]()
p(x) = ax2 + bx + c
∴ p(x) = (3k)x2 + (–k)x + ![]()
∴ p(x) = k (3x2 – x +
)
∴ p(x) =
(6x2 – 2x + 3) [Taking
common]
Couldn't generate an explanation.
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