Question

if alpha and beta are zeroes of a quadratic polynomial x^2 -5, then form a quadratic polynomial whose zeroes are 1+alpha and 1+beta. ​

Answer 1

2+√20/2 it is 1+ alpha

2-√20/2 it is 1+beta

Answer 2

Answer:

Required polynomial =

= x^2 - 2x - 4

Step-by-step explanation:

Given,

p(x) = x^2 - 5

Let the zeroes of this given polynomial be ' alpha ' and 'beta'.

Then,

x^2 - 5 = 0

x^2 = 5

x = √5

For the values of x,

CASE I -

x = + √5 (alpha)

CASE II -

x = - √5 (beta)

Hence,

x = +√5 , -√5

NOW,

Let the zeroes of new required polynomial be (1+alpha) and (1+beta).

Let the required polynomial be

ax^2 + bx + c

We know that,

(1+alpha) + (1+beta) = -b/a

=> (1+√5) + (1 +(-√5)) = -b/a

=> 1 + √5 + 1 -√5 = -b/a

=> 2 = -b/a

=> 2/1 = -b/a ...(1)

ALSO,

(1+alpha)(1+beta) = c/a

=> (1+√5)(1+(-√5)) = c/a

=> ( 1+√5)(1-√5) = c/a

=> 1 -√5 +√5 -5 = c/a

=> -4 = c/a

=> -4/1 = c/a ...(2)

On comparing both sides of eq.(1) and eq.(2) , we get

a = 1 , b = -2 , c = -4

So the required polynomial =>

= x^2 - 2x - 4