Question

if alpha and beta are zeros of a polynomial x square - 2 x-15 then from a quadratic polynomial whose zeros are 2 alpha and 2 Beta​

Answer 1

$\sf\blue{Question}$

$\sf{If \ \alpha \ and \ \beta \ are \ zeroes \ of \ a}$

$\sf{polynomial \ x^{2}-2x-15, \ then \ form \ a \ quadratic}$

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

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

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

$\sf{x^{2}-4x-60.}$

$\sf\orange{Given:}$

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

$\sf{\implies{x^{2}-2x-15}}$

$\sf{\implies{Zeroes \ are \ \alpha \ and \ \beta.}}$

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

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

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

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

$\sf{\implies{x^{2}-2x-15}}$

$\sf{Here \ a=1, \ b=-2 \ and \ c=-15.}$

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

$\sf{\therefore{\alpha+\beta=2...(1)}}$

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

$\sf{\therefore{\alpha\beta=-15...(2)}}$

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$\sf{Let \ M \ and \ N \ be \ the \ zeroes \ of \ required}$

$\sf{polynomial.}$

$\sf{\therefore{M=2\alpha \ and \ N=2\beta}}$

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

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

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

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

$\sf{\therefore{M+N=4...(3)}}$

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

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

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

$\sf{\therefore{M\times \ N=4(-15)}}$

$\sf{\therefore{M\times \ N=-60...(4)}}$

$\sf{Quadratic \ polynomial \ is}$

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

$\sf{\implies{x^{2}-(M+N)x+(M\times \ N)}}$

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

$\sf{\implies{x^{2}-4x-60}}$

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

$\sf\purple{\tt{x^{2}-4x-60.}}$