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Question
In the following determine the set of values of k for which the given quadratic equation has real roots:
4x2 - 3kx + 1 = 0
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Solution
The given quadric equation is 4x2 - 3kx + 1 = 0, and roots are real
Then find the value of k.
Here, a = 4, b = -3k and c = 1
As we know that D = b2 - 4ac
Putting the value of a = 4, b = -3k and c = 1
= (-3k)2 - 4 x (4) x (1)
= 9k2 - 16
The given equation will have real roots, if D ≥ 0
⇒ 9k2 - 16 ≥ 0
⇒ 9k2 ≥ 16
⇒ k2 ≥ 16/9
`rArrk>=sqrt(16/9)` or `k<=-sqrt(16/9)`
⇒ k ≥ 4/3 Or k ≤ -4/3
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