Question

If the equation x² + 4x + k = 0 has real and distinct roots, then
(a)k < 4
(b)k > 4
(c)k ≥ 4
(d)k ≤ 4

Answer 1

SOLUTION :  

Option (a) is correct : k < 4

Given : x² + 4x + k = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1 , b = 4  , c = k

D(discriminant) = b² – 4ac

D = (4)² - 4 × 1 × k

D = 16 - 4k  

D  > 0 (Given : real and distinct roots)

16 - 4k   > 0

-4k  > - 16

k < - 16/- 4

[When we multiply or divide by a negative number, We reverse the inequality]

k < 4

Hence, the value of k is < 4

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

options a will be right