Find the values of k for equation have real rootskx(x – 3) + 9 = 0

1.	Find the values of k for equation have real rootskx(x – 3) + 9 = 0
Answer» Given, kx(x – 3) + 9 = 0 It can be rewritten as, kx² – 3kx + 9 = 0 It’s of the form of ax²+ bx + c = 0 Where, a = k, b = -3k, c = 9 For the given quadratic equation to have real roots D = b²– 4ac ≥ 0 D = (-3k)² – 4(k)(9) ≥ 0 ⇒ 9k² – 36k ≥ 0 ⇒ 9k(k – 4) ≥ 0 ⇒ k ≥ 0 and k ≥ 4 ⇒ k ≥ 4 The value of k should be greater than or equal to 4 to have real roots.

Answer»

Given,

kx(x – 3) + 9 = 0

It can be rewritten as,

kx² – 3kx + 9 = 0

It’s of the form of ax²+ bx + c = 0

Where, a = k, b = -3k, c = 9

For the given quadratic equation to have real roots D = b²– 4ac ≥ 0

D = (-3k)² – 4(k)(9) ≥ 0

⇒ 9k² – 36k ≥ 0

⇒ 9k(k – 4) ≥ 0

⇒ k ≥ 0 and k ≥ 4

⇒ k ≥ 4

The value of k should be greater than or equal to 4 to have real roots.

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