Question

If alpha and beta are the zeros of 2x^2_5x+7 find a polynomial whose zeros are 2alpha+3beta and 3alpha+2beta

Answer 1

ANSWER:

Given

α & β are the roots of the polynomial 2x² - 5x + 7

p(x) = 2x² - 5x + 7

Comparing with ax² - bx + c

a = 2 , b = 5 , c = 7

We know that

Sum of roots (α + β) = - b/a

α + β = - (- 5) /2 = 5/2

Product of roots (αβ) = c/a

αβ = 7/2

Given 2α + 3β , 3α + 2β are the roots of the required quadratic polynomial

Quadratic polynomial = x² - (sum of roots)x + product of roots

Sum of roots of required polynomial = 2α + 3β+ 3α + 2β = 5(α + β)

Given α + β = 5/2

5(α + β) = 5(5/2) = 25/2

Product of roots

(2α + 3β)(3α + 2β) = 6α² + 4αβ + 9αβ + 6β²

= 6(α² + β²) + 13αβ

Using

a² + b² = (a + b)² - 2ab

Product of roots = 6[(α + β)² - 2αβ] + 13αβ

Product of roots = 6[(5/2)² - 2(7/2)] + 13(7/2)

Product of roots = 6[(25/4) - 7] + 91/2

Product of roots = 6(-3/4) + 91/2 = - 9/2 + 91/2 = 41

Now,

Quadratic polynomial = x² - (sum of roots)x + product of roots

= x² - (25/2)x + 41

= 2x² - 25x + 82

Hence , required polynomial is 2x² - 25x + 82