if alpha and beta are zeroes of polynomial p(x)=2x2-7x+5. Find a polynomial whose zeroes are 2alpha+1 and 2beta+3

Answer :

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

Equation  f ( x ) = 2 x²- 7 x  + 5
So,
We know from relationship between zeros and coefficient ,

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

$\alpha$ + $\beta$ =  $\frac{7}{2}$            --- ( 1 )
And

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

Taking Whole square of equation 1 , we get 

( $\alpha$ + $\beta$  )² =  $\frac{49}{4}$ 

And

( $\alpha$ -  $\beta$  )²  + 4$\alpha$$\beta$ =  $\frac{49}{4}$   , Substitute value from equation 2 , we get

( $\alpha$ -  $\beta$  )²  + 4 ( $\frac{5}{2}$ ) =  $\frac{49}{4}$ 

( $\alpha$ -  $\beta$  )²  + 10 =  $\frac{49}{4}$ 

( $\alpha$ -  $\beta$  )²  = $\frac{49}{4}$ - 10

( $\alpha$ -  $\beta$  )²  = $\frac{49 - 40}{4}$

( $\alpha$ -  $\beta$  )²  = $\frac{9}{4}$

$\alpha$ -  $\beta$   =  $\frac{3}{2}$                        ---- ( 3 )

Add equation 1 and 3 , We get

2$\alpha$   =   5

$\alpha$   =  $\frac{5}{2}$  , Substitute that value in equation 3 we get

$\frac{5}{2}$  - $\beta$ = $\frac{3}{2}$    , So

$\beta$  = 1

So,

We  get  $\alpha$   = $\frac{5}{2}$   and $\beta$  = 1

Then

2$\alpha$  + 1 = 2 ( $\frac{5}{2}$  ) + 1   =5 + 1  = 6
And
2  $\beta$  + 3 = 2 ( 1  ) + 3   = 2 + 3  = 5

Then

Sum of zeros of required polynomial  = (2$\alpha$  + 1 )+  (2  $\beta$  + 3 ) =  6 +  5 = 11
And

Product of zeros of required polynomial  = (2$\alpha$  + 1 )(2  $\beta$  + 3 )= ( 6 ) ( 5 ) = 30

And we know formula for polynomial when some of zeros and product of zeros we know :


Polynomial  =  k [ x²  - ( Sum of zeros ) x  + ( Product of zeros ) ]   , Here k is any non zero real number.

Substitute values , we get

Polynomial  =  k [ x²  - ( 11  ) x  + ( 30 ) ] 

Polynomial  =  k [ x²  - 11 x  + 30] 

Then

Quadratic polynomial  =    k [ x²  - 11 x  + 30]    =   x²  - 11 x  + 30    [on taking k = 1]                              ( Ans )