In the given polynomial x²-2x-√5 find the value of a) alpha-beta b)alpha/beta c) {(alpha)²-(beta)² - (alpha-beta)² - (alpha+beta)²} where alpha and beta are the zeroes of the polynomial.

Answer :

Given :$\alpha$ and  $\beta$   are two zeros of the polynomial

Equation  f ( x ) = x²- 2 x  - $\sqrt{5}$
So,
We know from relationship between zeros and coefficient ,

Sum of zeros  = $-\frac{coefficient x}{coefficient x^2}$ , So

$\alpha$ + $\beta$ =  2            --- ( 1 )
And

Product of zeros  = $\frac{cons\tan t term}{coefficient x^2}$ , So
$\alpha$$\beta$  =  - $\sqrt{5}$               ------ ( 2 )

Taking Whole square of equation 1 , we get 

( $\alpha$ + $\beta$  )² =  2²

And

( $\alpha$ -  $\beta$  )²  + 4$\alpha$$\beta$ =  4   , Substitute value from equation 2 , we get

( $\alpha$ -  $\beta$  )²  + 4 ( - $\sqrt{5}$ ) =  4

( $\alpha$ -  $\beta$  )²  - 4$\sqrt{5}$=  4

( $\alpha$ -  $\beta$  )²  = 4 + 4$\sqrt{5}$

( $\alpha$ -  $\beta$  )²  = 4 (  1 + $\sqrt{5}$ )

$\alpha$ -  $\beta$   =  2 $\sqrt{1 + \sqrt{5}}$                       ---- ( 3 )

Add equation 1 and 3 , We get

2$\alpha$   =   2 ( 1 + $\sqrt{1 + \sqrt{5}}$  )

$\alpha$   =  ( 1 + $\sqrt{1 + \sqrt{5}}$  )  , Substitute that value in equation 1 we get

( 1 + $\sqrt{1 + \sqrt{5}}$  )  +  $\beta$ = 2 , So

$\beta$  =  2 -  ( 1 + $\sqrt{1 + \sqrt{5}}$  )

$\beta$  =  1 -  $\sqrt{1 + \sqrt{5}}$ 

So,

We  get  $\alpha$   = ( 1 + $\sqrt{1 + \sqrt{5}}$  )   and $\beta$  = 1 -  $\sqrt{1 + \sqrt{5}}$   

So,

a ) From equation 3 :


$\alpha$ -  $\beta$   =  2 $\sqrt{1 + \sqrt{5}}$                                         (  Ans )

b )
$$\begin{gathered} \frac{\alpha }{\beta } = \frac{1 + \sqrt{1 + \sqrt{5}}}{1 - \sqrt{1 + \sqrt{5}}} \\ \Rightarrow \frac{\alpha }{\beta } = \frac{1 + \sqrt{1 + \sqrt{5}}}{1 - \sqrt{1 + \sqrt{5}}} \times \frac{1 + \sqrt{1 + \sqrt{5}}}{1 + \sqrt{1 + \sqrt{5}}} \\ \Rightarrow \frac{\alpha }{\beta } = \frac{1 + 1 + \sqrt{5} + 2\sqrt{1 + \sqrt{5}}}{1 - 1 - \sqrt{5}} \\ \Rightarrow \frac{\alpha }{\beta } = \frac{2 + \sqrt{5} + 2\sqrt{1 + \sqrt{5}}}{- \sqrt{5}} \\ \Rightarrow \frac{\alpha }{\beta } = \frac{2 + \sqrt{5} + 2\sqrt{1 + \sqrt{5}}}{- \sqrt{5}} \times \frac{-\sqrt{5}}{-\sqrt{5}} \\ \Rightarrow \frac{\alpha }{\beta } = \frac{-2\sqrt{5} - 5 - 2\sqrt{5\left(1 + \sqrt{5}\right)}}{5} \\ \mathbf{\Rightarrow }\frac{\mathbf{\alpha }}{\mathbf{\beta }}\mathbf{ }\mathbf{=}\mathbf{-}\mathbf{ }\frac{\mathbf{2}\sqrt{\mathbf{5}}\mathbf{ }\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{5}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{2}\sqrt{\mathbf{5}\left(\mathbf{1}\mathbf{ }\mathbf{+}\mathbf{ }\sqrt{\mathbf{5}}\right)}}{\mathbf{5}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)} \end{gathered}$$

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