Question

Find the values of k for which the given quadratic equation has real and distinct roots:

(i) kx² + 6x + 1 = 0

(ii) x² - kx + 9 = 0

(iii) 9x² + 3kx + 4 = 0

(iv) 5x² - kx + 1 = 0

Answer 1

(i) The given equation is kx² + 6x + 1 = 0

∴ D = 6² -4 x k x 1 = 36 - 4k

The given equation has real and distinct roots if D > 0.

∴ 36 - 4k > 0

⇒ 4k, 36

⇒ k < 9

(ii) The given equation is x² - kx + 9 = 0

∴ D = (-k)² - 4 x 1 x 9 = k² - 36

The given equation has real and distinct roots if D > 0.

∴ k² - 36 > 0

⇒ (k - 6) (k + 6) > 0

⇒ k < -6 or k > 6

(iii) The given equation is 9x² + 3kx + 4 = 0

∴ D = (3k)² - 4 x 9 x 4 = 9k² - 144

The given equation has real and distinct roots if D > 0.

∴ 9k² - 144 > 0

⇒ 9(k² - 16) > 0

⇒ (k - 4) ( k + 4) > 0

⇒ k < -4 or k > 4

(iv) The given equation is 5x² - kx + 1 = 0

∴ D = (-k)² - 4 x5 x 1 = k² - 20

The given equation has real and distinct roots if D > 0

∴ k² - 20 > 0

⇒ k² - (2√5)² > 0

⇒ (k - 2√5)(k +  2√5) > 0

⇒ k < - 2√5 or k > 2√5