The roots of each of the following quadratic equations are real and eq

1.	The roots of each of the following quadratic equations are real and equal, find k. i. 3y2 + ky + 12 = 0 ii. kx (x-2) + 6 = 0
Answer» i. 3y² + kg + 12 = 0 Comparing the above equation with ay² + by + c = 0, we get a = 3, b = k, c = 12 ∴ ∆ = b² – 4ac = (k)² – 4 × 3 × 12 = k² – 144 = k² – (12)² ∴ ∆ = (k + 12) (k – 12) …[∵ a – b = (a + b) (a – b)] Since, the roots are real and equal. ∴ ∆ = 0 ∴ (k + 12) (k – 12) = 0 ∴ k + 12 = 0 or k – 12 = 0 ∴ k = -12 or k = 12 ii. kx (x – 2) + 6 = 0 ∴ kx² – 2kx + 6 = 0 Comparing the above equation with ax² + bx + c = 0, we get a = k, b = -2k, c = 6 ∴ ∆ = b – 4ac = (-2k)² – 4 × k × 6 = 4k² – 24k ∴ ∆ = 4k (k – 6) Since, the roots are real and equal. ∴ ∆ = 0 ∴ 4k (k – 6) = 0 ∴ k(k – 6) = 0 ∴ k = 0 or k – 6 = 0 But, if k = 0 then quadratic coefficient becomes zero. ∴ k ≠ 0 ∴ k = 6

The roots of each of the following quadratic equations are real and equal, find k. i. 3y2 + ky + 12 = 0 ii. kx (x-2) + 6 = 0

Answer»

i. 3y² + kg + 12 = 0

Comparing the above equation with ay² + by + c = 0, we get

a = 3, b = k, c = 12

∴ ∆ = b² – 4ac

= (k)² – 4 × 3 × 12

= k² – 144 = k² – (12)²

∴ ∆ = (k + 12) (k – 12) …[∵ a – b = (a + b) (a – b)]

Since, the roots are real and equal.

∴ ∆ = 0

∴ (k + 12) (k – 12) = 0

∴ k + 12 = 0 or k – 12 = 0

∴ k = -12 or k = 12

ii. kx (x – 2) + 6 = 0

∴ kx² – 2kx + 6 = 0

Comparing the above equation with ax² + bx + c = 0, we get

a = k, b = -2k, c = 6

∴ ∆ = b – 4ac

= (-2k)² – 4 × k × 6

= 4k² – 24k

∴ ∆ = 4k (k – 6)

Since, the roots are real and equal.

∴ ∆ = 0

∴ 4k (k – 6) = 0

∴ k(k – 6) = 0

∴ k = 0 or k – 6 = 0 But, if k = 0 then quadratic coefficient becomes zero.

∴ k ≠ 0

∴ k = 6