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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 the equation x26x+k=0 x^2 - 6x + k = 0 are real and distinct, then the value of k k is:

(a) > –9
(b) > –6
(c) < 6
(d) < 9

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Solution

The roots of the quadratic equation ax² + bx + c = 0 are real and distinct if the discriminant (b² - 4ac) is greater than zero.

In the given equation x² - 6x + k = 0, a = 1, b = -6, and c = k.

So, the discriminant is (-6)² - 41k = 36 - 4k.

For the roots to be real and distinct, 36 - 4k > 0.

Solving this inequality gives k < 9.

So, the correct option is (d) < 9.

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