Question

If alfa and Beta are the zeroes of the polynomial x2+4x+3, form the polynomial whose
zeroes are 1+ alfa/beta and 1+ alfa/beta .​

Answer 1

Question :- If alfa and Beta are the zeroes of the polynomial x²+ 4x + 3, form the polynomial whose zeroes are {1 + (alfa/beta)} and {1 + (beta/alfa)} . ?

Solution :-

we know that,

• The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

• Product of roots of the Equation is given by = c/a.

comparing the given equation x² + 4x + 3, with ax² + bx + c we get :-

• a = 1
• b = 4
• c = 3

So,

→ sum of roots = (-b/a)

→ ɑ + β = (-4/1) = (-4)

and,

→ Product of roots = c/a

→ ɑβ = (3/1) = 3.

______________

Now, we have to form a polynomial whose

zeroes are {1 + (ɑ/β)} and {1 + (β/ɑ)}.

So,

→ Sum of roots of Required Equation = {1 + (ɑ/β)} + {1 + (β/ɑ)}

→ 2 + (ɑ/β) + (β/ɑ)

→ 2 + [ ɑ² + β² /ɑβ ]

→ 2 + [ (ɑ + β)² - 2ɑβ / ɑβ ]

Putting values from above now, we get,

→ 2 + [(-4)² - 2*3 / 3 ]

→ 2 + [ (16 - 6) / 3 ]

→ 2 + (10/3)

→ (6 + 10)/3

→ (16/3).

and,

→ Sum of roots of Required Equation = {1 + (ɑ/β)} * {1 + (β/ɑ)}

→ 1 + (ɑ/β) + 1 + (β/ɑ)

→ 2 + (ɑ/β) + (β/ɑ)

Putting value from above ,

→ (16/3).

Therefore,

→ Required Polynomial = x² - (sum of roots)x + Product of roots = 0 .

→ Required Polynomial = x² - (16/3)x + (16/3) = 0

→ Required Polynomial = (3x² - 16x + 16) /3 = 0

→ Required Polynomial = 3x² - 16x + 16 = 0 . (Ans.)

Answer 2

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