Question

in the following, determine the set of values of k for which the given quadratic equation has real roots: kx^2 + 6x +1 = 0

Answer 1

Given:

The root of the equation are real .

Equation -: kx² + 6x +1 = 0

To find :

Value of k

Solution :

kx² + 6x +1 = 0

Here a = 1 , b = 6 , c = 1

⇢D = b² - 4ac

⇢D = 6² - 4 × k × 1

⇢D = 36 - 4k

For real distinct roots , D ≥ 0

So , For real distinct root ,

⇢ 0 ≥ 36 - 4k

⇢ 0 - 36 ≥ -4k

⇢ -36 ≥ -4k

⇢ 4k ≤ 36

⇢k ≤ 36/4

⇢ k ≤ 9

Hence , k must be smaller or equal to 9 so that the equation has real roots .