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If the roots of $5 x^{2} - kx + 1 = 0$ are real and distinct then

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- 2 \sqrt{5} < k < 2 \sqrt{5}

k > 2 \sqrt{5}

only

k < - 2 \sqrt{5}

only

either

k > 2 \sqrt{5} or k < - 2 \sqrt{5}

answer is D.

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Detailed Solution

Given quadratic equation is,

5 x^{2} - kx + 1 = 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 and distinct roots if

b^{2} - 4 ac > 0

.
After comparing the given equation

5 x^{2} - kx + 1 = 0

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

a = 5, b = - k, c = 1

.
Then,

⟹ b^{2} - 4 ac > 0

⟹ {(- k)}^{2} - 4 \times 5 \times 1 > 0

⟹ k^{2} - 20 > 0

⟹ k^{2} > 20

⟹ k > 2 \sqrt{5} or k < - 2 \sqrt{5}

∴

If the roots of

5 x^{2} - kx + 1 = 0

are real and distinct then either

k > 2 \sqrt{5} or k < - 2 \sqrt{5}

.
Hence, the correct option is 4.

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