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Question
In the following determine the set of values of k for which the given quadratic equation has real roots:
x2 - kx + 9 = 0
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Solution
The given quadric equation is x2 - kx + 9 = 0, and roots are real.
Then find the value of k.
Here, a = 1, b = -k and c = 9
As we know that D = b2 - 4ac
Putting the value of a = 1, b = -k and c = 9
= (-k)2 - 4 x (1) x (9)
= k2 - 36
The given equation will have real roots, if D ≥ 0
⇒ k2 - 36 ≥ 0
⇒ k2 ≥ 36
`rArrk>=sqrt36` Or `k<=sqrt36`
⇒ k ≥ 6 Or k ≤ -6
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