Question

In the following, determine the set of values of k for which the given quadratic equation has real roots:

(i) 2x² + 3x + k = 0

(ii) 2x² + kx + 3 = 0

(iii) 2x² - 5x - k = 0

(iv) kx² + 6x + 1 = 0

(v) x² - kx + 9 = 0

Answer 1

(i) 2x² + 3x + k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

2x² + 3x + k = 0

⇒ D = 9 – 4 × 2 × k 

⇒ 9 – 8k ≥ 0 

⇒ k ≤ 9/8

(ii) 2x² + kx + 3 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

2x² + kx + 3 = 0

⇒ D = k² – 4 × 2 × 3

D ≥ 0

⇒ k² – 24 ≥ 0

⇒ (k + 2√6)(k – 2√6) ≥ 0

Thus, k ≤ - 2√6 or k ≥ 2√6

(iii) 2x² - 5x - k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

2x² - 5x - k = 0

⇒ D = 25 – 8k

D ≥ 0

⇒ 25 – 8k ≥ 0 

⇒ k ≤ 25/8

(iv) kx² + 6x + 1 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

kx² + 6x + 1 = 0

⇒ D = 36 – 4k 

⇒ 36 – 4k ≥ 0 

⇒ k ≤ 9

(v) x² - kx + 9 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

x² - kx + 9 = 0

⇒ D = k² – 36 

⇒ k² – 36 ≥ 0 

⇒ (k – 6)(k + 6) ≥ 0 

⇒ k ≥ 6 or k ≤ - 6