Math, asked by kajalshah, 1 year ago

Find the polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

Answers

Answered by AditiHegde
11

Given:

The polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

To find:

Find the polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

Solution:

From given, we have,

The polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

x^2+x-50 = 0

a = 1, b = 1 and c = -50

sum of the roots = α + β = -b/a = -1

product of the roots = αβ = c/a = -50

The polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

(α + β)² = (-1)² = 1

(αβ)² = (-50)² = 2500

The quadratic equation,

x² - (sum of roots)x + (product of roots) = 0

x² - ((α + β)² )x + ((αβ)² ) = 0

x² - 1x + 2500 = 0

x² - x + 2500 = 0 is the required equation

Answered by rudrakshvashisth12
4

Answer:

x ^ 2 - x + 2500

Step-by-step explanation:

From given, we have,

The polynomial whose zeros are squares of the zeros of polynomial x^2+x-50

a=1.b = 1andc = - 50

x^ ^2 + x - 50 = 0 sum of the roots = a + ẞ = -b/a = -1 product of the roots = aẞ = c/a = -50

The polynomial whose zeros are squares of the zeros of polynomial x^ ^2 + x - 50

(alpha + beta) ^ 2 = (- 1) ^ 2 = 1

(alpha*beta) ^ 2 = (- 50) ^ 2 = 2500

The quadratic equation, x² - (sum of roots)x + (product of roots) =

x ^ 2 - ((alpha + beta) ^ 2) * x + ((alpha*beta) ^ 2) = 0

x ^ 2 - x + 2500 = 0

x ^ 2 - x + 2500 = 0 is the required equation

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