Question

if alpha and beta are the zeros of a polynomial x square - 1 then find the polynomial whose zeros are 2 alpha divided by beta and 2 Beta / Alpha

Answer 1

Hi Mate !!


Here's ur ans :-


Given equation :- x² - 1

Let's factorise it by using identity :-
a² - b² = ( a - b ) ( a + b )

x² - 1 = 0

x² - ( 1 )² = 0

( x + 1 ) ( x - 1 ) = 0


• ( x + 1 ) = 0

x = ( - 1 )

• ( x - 1 ) = 0

x = 1


So,
$let \: \alpha \: be \: ( - 1) \: \: and \: \: \beta \: \: be \: \: 1$


# For new equation having Zeros as :-

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

$= ) \frac{2 \alpha }{ \beta } = \frac{2 \times ( - 1)}{1} = ( - 2)$


$= ) \frac{2 \beta }{ \alpha } = \frac{2 \times 1}{ -1} = \frac{2}{ - 1} = - 2$


So, the Zeros of new equation are ( - 2 ) and ( - 2 )

• Sum of the Zeros

( - 2 ) + ( - 2 )

- 2 - 2

= ( - 4 )


• Product of Zeros :-

( - 2 ) × ( - 2 )

= 4



♯ To form the quadratic equation we have formula as :-

x² - ( Sum of Zeros )x +(Product of Zeros )


Putting value in the formula :+


x² - ( - 4 )x + ( 4 ) = 0

x² + 4x + 4 = 0

So, the required quadratic equation is :-
x² + 4x + 4 = 0

Answer 2

I am taking

alpha =A & beta =B

since,

f(x) = x^2 - 1

A + B = -b/a

= -(0) /1

=0/1

= 0

A × B = c/a

= -1/1

= 1

Therefore,

A+B=0 & A×B=-1

Since,

A=2A/B, B=2B/A are the two zeros

=> sum of the given zeroes is

2A/B + 2B/A = (2A^2+2B^2) /AB

And

Product of the two zeroes is

( 2A/B) ×(2B/A) = 4

Since,

f(x) =k[x^2-(A+B) x +AB]

f(x) =x^2-(2A/B + 2B/A) x + (2A/B) (2B/A)

f(x) =x^2-[(2A^2 + 2B^2) /-1] x + 4

f(x) =x^2-[2(A^2+B^2) /-1]x + 4

f(x) =x^2-[2{(A+B) ^2-2AB}/-1]x + 4

f(x) =x^2-[2(0) ^2-2(-1) /-1]x + 4

f(x) =x^2-(2×0+2/-1) x + 4

f(x) =x^2-(0+2/-1) x + 4

f(x) =x^2-(2/-1) x + 4

f(x) =x^2-(-2) x + 4

f(x) =x^2 + 2x + 4

Therefore the equation is

f(x) =x^2 + 2x + 4

Hope it's helpful

If any doubts or queries please do ask me