Question

Find the quadratic polynomial for the zeroes 1/3,-2.

Answer 1

$\huge\mathfrak{Solution :}$

Given zeroes :-

1 / 3 and -2

Let

$\alpha = \frac{1}{3 }$

And

$\beta = - 2$

Now,

Sum of the zeroes :-

$= \alpha + \beta \\ \\ \\ = \frac{1}{3} + ( - 2) \\ \\ \\ = \frac{1 - 6}{3} \\ \\ \\ = \frac{ - 5}{3}$

Product of zeroes :-

$= \alpha \beta \\ \\ \\ = \frac{1}{3} \times - 2 \\ \\ \\ = \frac{ - 2}{3}$

Now,

Required polynomial is :-

p(x) = x² - (Sum of zeroes)x + Product of zeroes

$p(x) = {x}^{2} - ( \frac{ - 5}{3} ) + ( \frac{ - 2}{3} ) \\ \\ \\ p(x) = {x}^{2} + \frac{5}{3} - \frac{2}{3}$

$\huge\mathbb{Hope \ this \ helps.}$

Answer 2

Sum of the zeros=alpha+beta
=1/3+(-2)
=-5/3
Product of the zeros=alpha*beta
=1/3*(-2)
=-2/3
The required quadratic polynomial=x^2-(alpha+beta)x+(alpha.beta)
=x^2+5/3x-2/3
or
=3x^2+5x-2
Hope you satisfied..
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