Question

if Alfa and beta are the zero of the polynomial 3x^+5x-2 then form a quardiatic polynomial whose zero are 2alfa and 2 beta​

Answer 1

$\sf\blue{Correct \ Question}$

$\sf{If \ \alpha \ and \ \beta \ are \ the \ zero}$

$\sf{of \ the \ polynomial \ 3x^{2}+5x-2 \ then}$

$\sf{form \ a \ quadratic \ polynomial \ whoes}$

$\sf{zeroes \ are \ 2\alpha \ and \ 2\beta.}$

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$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ quadratic \ polynomial \ with \ required}$

$\sf{zeroes \ is \ 3x^{2}+10x-8.}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{3x^{2}+5x-2}}$

$\sf{\implies{Zeroes \ of \ the \ polynomial \ are \ \alpha \ and \ \beta}}$

$\sf\pink{To \ find:}$

$\sf{The \ quadratic \ polynomial \ whose}$

$\sf{zeroes \ are \ 2\alpha \ and \ 2\beta.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{3x^{2}+5x-2}}$

$\sf{Here, \ a=3, \ b=5 \ and \ c=-2}$

$\sf{Sum \ of \ zeroes=\frac{-b}{a}}$

$\sf{\therefore{\alpha+\beta=\frac{-5}{3}...(1)}}$

$\sf{Product \ of \ zeroes=\frac{c}{a}}$

$\sf{\therefore{\alpha\beta=\frac{-2}{3}...(2)}}$

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$\sf{Let \ the \ zeroes \ of \ the \ required }$

$\sf{quadratic \ polynomial \ be \ M \ and \ N.}$

$\sf{Let \ M \ be \ 2\alpha \ and \ N \ be \ 2\beta}$

$\sf{M+N=2\alpha+2\beta}$

$\sf{M+N=2(\alpha+\beta)}$

$\sf{...from \ (1)}$

$\sf{M+N=2\times(\frac{-5}{3})}$

$\sf{\therefore{M+N=\frac{-10}{3}...(3)}}$

$\sf{M\times \ N=2\alpha\times2\beta}$

$\sf{M\times \ N=4\alpha\beta}$

$\sf{...from \ (2)}$

$\sf{M\times \ N=4\times(\frac{-2}{3})}$

$\sf{\therefore{M\times \ N=\frac{-8}{3}...(4)}}$

$\sf{Quadratic \ polynomial \ is}$

$\sf{x^{2}-(Sum \ of \ zeroes)x+(Product \ of \ zeroes)}$

$\sf{...from \ (3) \ and \ (4)}$

$\sf{\implies{x^{2}+\frac{10x}{3}-\frac{8}{3}}}$

$\sf{Multiply \ throughout \ by \ 3}$

$\sf{\implies{3x^{2}+10x-8}}$

$\sf\purple{\tt{\therefore{The \ quadratic \ polynomial \ with \ required}}}$

$\sf\purple{\tt{zeroes \ is \ 3x^{2}+10x-8.}}$

Answer 2

$\huge\underline\mathfrak\red{Question:}$

$if \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: the \: polynomial \\ \: 3 {x}^{2} + 5x - 2 \: then \: form \: a \: quadratic \\ polynomial \: whose \: zeros \: are \: 2 \alpha \: and \: 2 \beta \:$

$\huge\underline\mathfrak\blue{Answer:}$

$the \: quadratic \: polynomial \: is \\ \: 3 {x}^{2} + 10x - 8$

$\huge\underline\mathfrak\green{solution}$

$given \: a \: quadratic \: polynomial$$a {x}^{2} + bx + c$ $sum \: of \: its \: zeroes \:  \alpha \: and \beta \: is \ : - \frac{b}{a} \: and \\ product \: of \: zeroes \: is  \frac{c}{a}$

$further \: if \: sum \: of \: zeroes \: is  \: s \: and \: \\ product \: of \: zeroes \: is  \: p \: then \: quadratic \: polynomial \: is  \\ \: {x}^{2} - sx + p$

$hence \: for \: \: 3 {x}^{2} + 5x + 2 \: we \: have$

$\alpha + \beta = - \frac{5}{3}$

$and \\ \alpha \beta = - \frac{2}{3}$

$we \: now \: desire \: quadratic \: polynomial \\ whose \: zeroes \: are \:  2 \alpha \: and \:  2 \beta$

$and \: product \: of \: roots \: is \: \\ \\ 2 \alpha \times 2 \beta = 4 \alpha \beta = 4 \times - \frac{(2}{3} = - \frac{8}{3}$

$and \: quadratic \: polynomial \: is \: \\ \\ {x}^{2} + \frac{10}{3} x - \frac{8}{3}$

$as \: zeroes \: are \: not \: affected \: by \: multiplying \\ each \: term \: of \: polynomial \: by \: a \: constant,$

$we \: can \: say \: quadratic \: polynomial \: is \\ \\ 3 {x}^{2} + 10x - 8$

hope it helps:))