If one zero of the quadratic polynomial f(x) = 4x² – 8kx + 8x – 9 is negative of the other then Find the zeros of kx² + 3kx + 2.

f(x) = 4x² – 8kx + 8x – 9

f(x) = 4x² – 8(k – 1)x – 9

let one zero of f(x) be α

as other is negative hence –α

comparing 4x² – 8(k – 1)x – 9 with standard from ax² + bx + c = 0

a = 4, b = –8(k – 1) and c = –9

sum of roots

⇒ b = 0

⇒ –8(k – 1) = 0

⇒ k = 1

Now substituting k =1 in the quadratic polynomial kx² + 3kx + 2 we get x² + 3x + 2

Now let us Find its zeroes

⇒ x² + 3x + 2 = 0

⇒ x² + 2x + x + 2 = 0

⇒ x(x + 2) + 1(x + 2) = 0

⇒ (x + 1)(x + 2) = 0

⇒ x + 1 = 0 and x + 2 = 0

⇒ x = –1 and x = –2

Hence zeroes of kx² + 3kx + 2 as k = 1 are x = –1 and x = –2