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For what values of k, does the equation $k x^{2} - 6 x - 2 = 0$ have real roots?

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k \leq - \frac{9}{2}

k \geq - \frac{9}{2}

k \leq - 2

None of these

answer is B.

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Detailed Solution

Given quadratic equation is,

k x^{2} - 6 x - 2 = 0

A quadratic equation in the variable x is of the form

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers and has real roots if

b^{2} - 4 ac \geq 0

.
After comparing the given equation

k x^{2} - 6 x - 2 = 0

with the standard form of the quadratic equation, we get,

a = k, b = - 6, c = - 2

.
Then,

⟹ b^{2} - 4 ac \geq 0

⟹ {(- 6)}^{2} - 4 \times k \times - 2 \geq 0

⟹ 36 + 8 k \geq 0

⟹ - 8 k \geq 36

⟹ k \geq - \frac{36}{8}

⟹ k \geq - \frac{9}{2}

∴

For

k \geq - \frac{9}{2}

, the equation

k x^{2} - 6 x - 2 = 0

has real roots.
Hence, the correct option is 2.

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