Question

α and β are zeroes of a quadratic polynomial x² − ax − b. Obtain a quadratic polynomial whose zeroes are 3α + 1 and 3β + 1.

Answer 1

x² − ax − b

α + β = `(-(-a))/1`

= a

α . β = `(-b)/1`

= −b

Now, quadratic equation whose zeroes are 3α + 1 and 3β + 1 is

⇒ x² − (3α + 1 + 3β + 1)x + (3α + 1)(3β + 1)

⇒ x² − [3(α + β) + 2]x + 9αβ + 3(α + β) + 1

⇒ x² − [3a + 2]x + 9(−b) + 3a + 1

⇒ x² − (3a + 2)x − 9b + За + 1