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

kx^{2} + 6x + 1 = 0

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#### Solution

The given quadric equation is kx^{2} + 6x + 1 = 0, and roots are real

Then find the value of *k.*

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

As we know that D = b^{2} - 4ac

Putting the value of a = k, b = 6 and c = 1

= 6^{2} - 4 x (k) x (1)

= 36 - 4k

The given equation will have real roots, if D ≥ 0

36 - 4k ≥ 0

4k ≤ 36

k ≤ 36/4

k ≤ 9

Therefore, the value of k ≤ 9.

Concept: Nature of Roots

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