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