Question

If one zero of the polynomial ( k+1 ) x² - 5x+ 5 is multiplicative inverse of other , find the zeroes of kx² - 3kx +9 where I is constant

Answer 1

Answer:

3/2, 3/2

Explanation:

for (k+1)x² - 5x + 5

let one root be 'x'

⇒ other root will be '1/x'

a = (k+1); b = -5; c = 5

product of roots = c/a

⇒ (x)(1/x) = 5/(k+1)

⇒ 1 =  5/(k+1)

⇒ k+1 = 5

⇒k = 4

kx² - 3kx +9

⇒ (4)x² - 3(4)x + 9

⇒ 4x² - 12x + 9

⇒ (2x)² - 2(2x)(3) + (3)²

⇒ (2x - 3)²      [∵ (a-b)² = a² - 2ab +b²]

zeros of 4x² - 12x + 9 was

4x² - 12x + 9 = 0

⇒ (2x - 3)² = 0

⇒ x = 3/2

∴ the roots were 3/2, 3/2