Question

Find quadratic polynomial with sum of zeroes equal to 1/4 and product equal to 1/3

Answer 1

Given :

$\sf \implies Sum \: of \: zeroes \: ( \alpha + \beta) = \frac{1}{4} \\ \\ \sf \implies product \: of \: zeroes \: ( \alpha \beta) = \frac{1}{3}$

To Find :

$\implies$ Quadratic polynomial

Solution :

Formula used

$\large\implies \boxed{\boxed{ \sf{x}^{2} + ( \alpha + \beta)x + \alpha \beta}}$

Substitute values in formula

$\sf \implies {x}^{2} + \bigg( \frac{1}{4} \bigg )x + \frac{1}{3} = 0 \\ \\ \sf \implies {x}^{2} +\frac{x}{4}+ \frac{1}{3} = 0 \\ \\ \sf \implies \frac{12 {x}^{2} + 3x + 4}{12} = 0 \\ \\ \large\implies \boxed{ \boxed{ \sf12 {x}^{2} + 3x + 4 = 0}}$

Answer 2

✯✯ QUESTION ✯✯

Find quadratic polynomial with sum of zeroes equal to $\dfrac{1}{4}$and product equal to $\dfrac{1}{3}$..

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✰✰ ANSWER ✰✰

$\longrightarrow{Sum\:of\:Zeroes(\alpha+\beta)=\dfrac{1}{4}}$

$\longrightarrow{Product\:of\:Zeroes(\alpha\beta)=\dfrac{1}{3}}$

➜Using Formula : -

$\longrightarrow{\small{\boxed{\boxed{\bold{\purple{\sf{{x}^{2}+(\alpha+\beta)x+(\alpha\beta)}}}}}}}$

➜Putting Values : -

$\longrightarrow{{x}^{2}+\dfrac{1}{4}x+\dfrac{1}{3}=0}$

$\longrightarrow{{x}^{2}+\dfrac{1x}{4}+\dfrac{1}{3}=0}$

☛Taking L.C.M of Denominators : -

$\longrightarrow{\dfrac{12x^2+3x+4}{12}=0}$

➜Quadratic Polynomial is: - ${12x}^{2}+3x+4$