Question

If α and β are the zeroes of the polynomial x2−3x−2, find the quadratic polynomial whose zeroes are 12α+β and 12β+α.

A. 15x²−9x+2
B. 16x²+9x−1
C. 16x²−9x+1
D. none

Answer 1

Using Vieta's formulae and the given roots α and β from the polynomial x2−3x−2, we calculate the sum and product of the new roots 12α+β and 12β+α, leading to the quadratic polynomial x2 − 39x − 338 which is not listed, making option D correct.

Step-by-step explanation:

If α and β are the zeroes of the polynomial x2−3x−2, we apply Vieta's formulae which state that for any quadratic polynomial ax2+bx+c, the sum of its roots α and β is −b/a and the product is c/a. Given the polynomial x2−3x−2, we have α + β = 3 and αβ = −2. We wish to find the quadratic polynomial with roots 12α+β and 12β+α. Using the relationships between roots and coefficients, the sum of these new roots is (12α+β)+(12β+α) = 13α + 13β = 13(α + β), which is 13×3 = 39. The product of the new roots is (12α+β)(12β+α) = 12α×12β + α×12β + β×12α + αβ = 144αβ + 12αβ + 12αβ + αβ = 144αβ + 24αβ + αβ = 169αβ = 169(−2) = −338. Thus, the polynomial we seek is x2 − 39x − 338, which is not amongst the given options, so the correct choice is D. none.