Question

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

(i) 2x² + kx + 2 = 0

(ii) 3x² + 2x + k = 0

(iii) 4x² - 3kx + 1 = 0

(iv) 2x² + kx - 4 = 0

Answer 1

(i) 2x² + kx + 2 = 0

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

D = b² – 4ac

If D ≥ 0, roots are real

2x² + kx + 2 = 0

⇒ D = k² – 4 × 4 

⇒ k² – 16 ≥ 0 

⇒ (k + 4)(k – 4) ≥ 0 

⇒ k ≥ 4 or k ≤ -4

(ii) 3x² + 2x + k = 0

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

D = b² – 4ac

If D ≥ 0, roots are real

3x² + 2x + k = 0

⇒ D = 4 – 12k 

⇒ 4 – 12k ≥ 0 

⇒ k ≤ 1/3

(iii) 4x2 - 3kx + 1 = 0

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

D = b² – 4ac

If D ≥ 0, roots are real

4x² - 3kx + 1 = 0

⇒ D = 9k² – 16 

⇒ 9k² – 16 ≥ 0 

⇒ (3k – 4)(3k + 4) ≥ 0 

⇒ k ≤ -4/3 or k ≥ (4/3)

(iv) 2x² + kx - 4 = 0

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

D = b² – 4ac

If D ≥ 0, roots are real

2x² + kx - 4 = 0

⇒ D = k² + 4 × 2 × 4 = k² + 32

Thus, D is always greater than 0 for all values of k.