Question

frame a quadratic polynomial whose sum and product are -3 and 2​

Answer 1

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Sum of zeros (\alpha\:+\beta) = a

Product of zeros (\alpha\beta) = \dfrac{1}{a}

_______________ [GIVEN]

• We have to find the quadratic equation.

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» Sum of zeros = (\alpha\:+\beta) = a

» Product of zeros = \alpha\beta = \dfrac{1}{a}

We know that..

x² - (Sum of zeros)x + Product of zeros = 0

OR

x² - (\alpha\:+\beta)x + \alpha\beta = 0

\implies x² - (a)x + \dfrac{1}{a} = 0

\implies \dfrac{a {x}^{2} \: - \: {a}^{2}x \: + \: 1}{a} = 0

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ax² - a²x + 1 is the quadratic polynomial.

Answer 2

Let A and B be the roots of the quadratic polynomial.

Then, A+B= -3

          AB= 2

       

The required polynomial is X^2 - (A+B)X + AB

                                         => X^2 - (-3)X + 2

                                         => X^2 + 3X + 2