Question

find a quadratic polynomial whose zeroes are the reciprocal of the zeroes of 4xsquare -3x -1​

Answer 1

hope it helps you to find out the correct answer

Answer 2

Answer:-

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

$\sf{x^{2}+3x-4.}$

Given:

• The given polynomial is $\sf{4x^{2}-3x-1}$

To find:

• To find a quadratic polynomial whose are reciprocal of the zeroes of the given polynomial.

Solution:

$\sf{Let \ the \ zeroes \ of \ the \ polynomial \ be \ \alpha \ and \ \beta}$

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

$\sf{\implies{4x^{2}-3x-1}}$

$\sf{Here, \ a=4, \ b=-3 \ and \ c=-1}$

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

$\sf{\therefore{\alpha+\beta=\frac{3}{4}...(1)}}$

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

$\sf{\therefore{\alpha\beta=\frac{-1}{4}...(2)}}$

___________________________________

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

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

$\sf{\therefore{M=\frac{1}{\alpha} \ and \ N=\frac{1}{\beta}}}$

$\sf{M+N=\frac{1}{\alpha}+\frac{1}{\beta}}$

$\sf{\therefore{M+N=\frac{\alpha+\beta}{\alpha\beta}}}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\therefore{M+N=\frac{\frac{3}{4}}{\frac{-1}{4}}}}$

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

$\sf{M\times \ N=\frac{1}{\alpha}\times\frac{1}{\beta}}$

$\sf{\therefore{M\times \ N=\frac{1}{\alpha\beta}}}$

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

$\sf{\therefore{M\times \ N=\frac{1}{\frac{-1}{4}}}}$

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

$\sf{Quadratic \ polynomial \ is}$

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

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

$\sf{\implies{x^{2}-(-3)x+(-4)}}$

$\sf{\implies{x^{2}+3x-4}}$

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

$\sf\purple{\tt{x^{2}+3x-4.}}$