Question

a quadratic polynomial x^2 -2x-8 has the zeroes alph and beta. form a quadratic polynomial whose zeroes are 2/alpa 2/beta​

Answer 1

GIVEN :-

p(x) = x^2 - 2x - 8

$zeros \: are \: \alpha \: \: and \: \: \beta$

To Form :- A quadratic polynomial having

$\frac{2}{ \alpha } and \: \frac{2}{ \beta } as \: zeros$

alpha + beta = -b/a

aplha × beta = c/a

SOLUTION :-

First of all we haveto find the sum and product of new zeros of the polynomial.

A/Q :-

Sum of the zeros of new polynomial =

$\frac{2}{ \alpha } + \frac{2}{ \beta } \\ \\ \\ = \frac{2 \beta + 2 \alpha }{ \alpha \beta } \\ \\ \\ = \frac{2( \alpha + \beta )}{ \alpha \beta } \\ \\ \\ = \frac{2(\frac{ - b}{a} )}{ \frac{c}{a} } \\ \\ \\ = \frac{2( \frac{2}{1} )}{ \frac{ - 8}{1} } \\ \\ = \frac{4}{ - 8} = \frac{ - 1}{2}$

Product of the zeros of the new polynomial =

$\alpha \beta \\ \\ = \frac{2}{ \alpha } \times \frac{2}{ \beta } \\ \\ = \frac{4}{ \alpha \beta } \\ \\ = \frac{4}{ \frac{c}{a} } \\ \\ = \frac{4}{ \frac{ - 8}{1} } \\ \\ = \frac{ - 1}{2}$

NEW POLYNOMIAL =

$f(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ \\ = {x}^{2} - ( \frac{ - 1}{2} )x + \frac{ - 1}{2} \\ \\ = {x}^{2} + \frac{1}{2} x - \frac{1}{2} \\ \\ = 2 {x}^{2} + x - 1$

[ANSWER ]