Question

18. Find a quadratic polynomial whose zeroes are:-
$\frac{2}{3} \: and \: - \frac{1}{3}$
​

Answer 1

Answer:

9x^2 - 3x - 2 = 0

Step-by-step explanation:

P(x) = x^2 - (sum of zeroes/roots)x + (product of roots)

P(x) = x^2 - (2/3 -1/3)x + (2/3 X -1/3) = 0

Answer 2

$\Large{\underline{\underline{\mathfrak{\orange{\sf{Question}}}}}}$.

Find a quadratic polynomial whose zeroes are:-

$\frac{2}{3} \: and \: - \frac{1}{3}$

$\Large{\underline{\underline{\mathfrak{\orange{\sf{Answer}}}}}}$.

Given

• Zeroes of polynomial are $\frac{2}{3} \: and \: - \frac{1}{3}$

Find

• Equation of polynomial

$\Large\underline{\underline{\mathfrak{\orange{\sf{Explanation}}}}}$.

$\bigstar\sf{\:Sum\:of\:zeroes\:=\:\dfrac{2}{3}+\dfrac{-1}{3}}$

$\mapsto\sf{\:Sum\:of\:zeroes\:=\:\dfrac{2-1}{3}}$

$\mapsto\pink{\sf{\:Sum\:of\:zeroes\:=\:\dfrac{1}{3}}}$

$\bigstar\sf{\:Product\:of\:zeroes\:=\:\dfrac{2}{3}\times\dfrac{-1}{3}}$

$\mapsto\pink{\sf{\:Product\:of\:zeroes\:=\:\dfrac{-2}{9}}}$

We Know,

$\green{\boxed{\boxed{\orange{\small{\sf{\:x^2-(Sum\:of\:zeroes)x+(product\:of\:zeroes)\:=\:0}}}}}}$

$\:\:\:\:\:\:\small\green{\sf{\:keep\:values}}$

$\mapsto\sf{\:x^2-(\dfrac{1}{3})x+(\dfrac{-2}{9})\:=\:0}$

$\mapsto\sf{\:9x^2-3x-2\:=\:0}$

$\Large\pink{\sf{\underbrace{Thus}}}$

$\LARGE\underline{\underline{\sf{\green{\:Required\:Equation}}}}$

$\mapsto\orange{\sf{\:9x^2-3x-2\:=\:0}}$