If alpha and beta are the zeros of the polynomial 2x^2-5x+7, then find the polynomial whose zeros are 2alpha+3beta and 3alpha+2beta
Answers
EXPLANATION.
α and β are the zeroes of the polynomial.
⇒ 2x² - 5x + 7.
As we know that,
Sum of the zeroes of quadratic polynomial.
⇒ α + β = - b/a.
⇒ α + β = - (-5)/2 = 5/2.
Products of the zeroes of quadratic polynomial.
⇒ αβ = c/a.
⇒ αβ = 7/2.
To find polynomial :
Whose zeroes are 2α + 3β and 3α + 2β.
Sum of the zeroes of new quadratic polynomial.
⇒ (2α + 3β) + (3α + 2β).
⇒ (5α + 5β).
⇒ 5(α + β) = 5(5/2) = (25)/(2).
⇒ (α + β)' = (25/2).
Products of the zeroes of new quadratic polynomial.
⇒ (2α + 3β)(3α + 2β).
⇒ 2α(3α + 2β) + 3β(3α + 2β).
⇒ 6α² + 4αβ + 9αβ + 6β².
⇒ 6α² + 6β² + 4αβ + 9αβ.
⇒ 6[α² + β²] + 13αβ.
As we know that,
Formula of :
⇒ (x² + y²) = (x + y)² - 2xy.
Using this formula in the equation, we get.
⇒ 6[(α + β)² - 2αβ] + 13αβ.
Put the values in the equation, we get.
⇒ 6[(5/2)² - 2(7/2)] + 13(7/2).
⇒ 6[25/4 - 7] + (91/2).
⇒ 6[(25 - 28)/4] + (91/2).
⇒ 6[- 3/4] + (91/2).
⇒ - 18/4 + 91/2.
⇒ - 9/2 + 91/2.
⇒ 82/2 = 41
⇒ (αβ)' = 41.
As we know that,
Formula of quadratic polynomial.
⇒ x² - (α + β)x + αβ.
Put the values in the equation, we get.
⇒ x² - (25/2)x + 41.