Question

if alpha and beta are zeroes of the poly. p(x)= x^2 - 2x - 8 , find the quadratic poly. whose zeroes are 2alpha/beta and 2beta/alpha

Answer 1

Answer:

x^2-5x+4=0

Step-by-step explanation:

Hope you will understand it

Answer 2

Let the zeroes of this polynomial be alpha and beta,

$sum \: of \: zeroes \: = \frac{ - b}{a} \\ product \: of \: zeroes \: = \frac{c}{a} \\ \alpha + \beta = \frac{ - ( - 2)}{1} \\ \alpha + \beta = 2 \: \: \: \: \: \: \: eq {}^{n} \: 1 {}^{st} \\ \alpha \beta = \frac{ - 8}{1} \\ \alpha \beta = - 8 \\ finding \: polynomial \: from \: zeroes \\ k( {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) \\ k( {x}^{2} - ( \frac{2 \alpha }{ \beta } + \frac{2 \beta }{ \alpha } ) x + ( \frac{2 \alpha }{ \beta } \times \frac{2 \beta }{ \alpha } ) \\ k( {x}^{2} - ( \frac{ {2 \alpha }^{2} + {2 \beta }^{2} }{ \alpha \beta } )x + (4)) \\ k( {x}^{2} - (\frac{2( { \alpha }^{2} + { \beta }^{2}) }{ \alpha \beta } )x + 4) \\ for \: finding \: the \: polynomial \: we \: \\ need \: the \: value \: of \: { \alpha }^{2} + { \beta }^{2} . \\ { \alpha }^{2} + { \beta }^{2} = ( { \alpha + \beta )}^{2} - 2 \alpha \beta \\ put \: the \: value \: of \: ( \alpha + \beta ) \: and \: \alpha \beta . \\ ( {2})^{2} - 2( - 8) \\ 4 + 16 = 20 = { \alpha }^{2} + { \beta }^{2}$

$put \: the \: value \: of \: { \alpha }^{2}+ { \beta }^{2} \: in \: the \: expression \: \\ {x}^{2} + ( \frac{2( { \alpha }^{2} + { \beta }^{2} ) }{ \alpha \beta } )x + 4 \\ {x}^{2} + ( \frac{-2(20)}{ - 8} )x + 4 \\ {x}^{2} + ( \frac{-40}{ - 8} )x + 4 \\ {x}^{2} +5x + 4$

This polynomial has zeroes, 2alpha/beta and 2beta/alpha.

It was a nice question.