Question

If alpha and beta are the zeroes of polynomial x2+13x+22 then find the polynomial whose zeroes are alpha /beta +2,beta/alpha-2

Answer 1

Solution :

The given polynomial is

P(x) = x² + 13x + 22

= x² + 11x + 2x + 22

= x(x + 11) + 2(x + 11)

= (x + 11) (x + 2)

So, the zeroes be x = - 11, - 2

Let, α = - 11 and β = - 2

We have to find the polynomial whose zeroes are (α/β + 2) and (β/α - 2)

Now, α/β + 2 = (- 11)/(- 2) + 2

        = 11/2 + 2 = 15/2

and β/α - 2 = (- 2)/(- 11) - 2

     = 2/11 - 2 = - 20/11

Thus, the required polynomial is

Q(x) = (x - 15/2) (x + 20/11)

= x² + (20/11 - 15/2)x - 300/22

= x² + {(40 - 165)/22}x - 300/22

= [22x² - 125x - 300]/2

i.e., Q(x) = 22x² - 125x - 300

Answer 2

Answer:

22x - 125x - 300

Step-by-step explanation:

Given polynomial -

    x² + 13x + 22 = 0

⇒ x² + 11x + 2x + 22 = 0

⇒ x ( x + 11 ) + 2 ( x + 11 ) = 0

⇒ ( x + 11 ) ( x + 2 ) = 0

∴ x = - 11 or x = - 2

Let α = - 11 and β = - 2

We have to find a polynomial whose zeroes are (α/β + 2 ) and ( β/α - 2 ).

For (α/β + 2 ),

⇒ - 11 / - 2 + 2

⇒ 11 / 2 + 2

⇒ 15 / 2

For ( β/α - 2 ).

⇒ - 2 / - 11 - 2

⇒ 2 / 11 - 2

⇒ - 20 / 11

Therefore, the required polynomial is -

P (x) = ( x - 15 / 2 ) ( x + 20 / 11 )

⇒ x² + (20 / 11 - 15 / 2)x - 15/ 2 * 20 / 11

⇒ x² + {( 40 - 165)/22}x - 300 / 22

⇒ 22x² - 125x - 300 / 22 = 0

⇒ 22x² - 125x - 300 = 0

Hence, the required polynomial is 22x² - 125x - 300.