If the roots of equation 𝑥2 − 6𝑥 + 𝑘 = 0 are real and distinct, then value of k is:(a) > –9(b) > –6(c) < 6(d) < 9
Question
If the roots of equation 𝑥2 − 6𝑥 + 𝑘 = 0 are real and distinct, then value of k is:(a) > –9(b) > –6(c) < 6(d) < 9
Solution
The roots of a quadratic equation ax² + bx + c = 0 are real and distinct if the discriminant (b² - 4ac) is greater than zero.
In this case, the equation is x² - 6x + k = 0. So, a = 1, b = -6, and c = k.
The discriminant is (-6)² - 41k = 36 - 4k.
For the roots to be real and distinct, the discriminant must be greater than zero.
So, 36 - 4k > 0.
Solving for k, we get k < 9.
So, the correct option is (d) < 9.
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