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Question
In the following determine the set of values of k for which the given quadratic equation has real roots:
kx2 + 6x + 1 = 0
Solution
The given quadric equation is kx2 + 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 = b2 - 4ac
Putting the value of a = k, b = 6 and c = 1
= 62 - 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.
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