If the equation $k x^{2} + 4 x + 1 = 0$ has real and distinct roots, then:

k < 4

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k > 4

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k ⩽ 4

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k ⩾ 4

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Solution

The correct option is A $k < 4$
Given: quadratic equation $k x^{2} + 4 x + 1 = 0$ , has real and distinct roots
To find the value of k
Sol: An equation has real and distinct roots if the discriminant $b^{2} - 4 a c > 0$
In the given equation, $a = k, b = 4, c = 1$
Therefore the discriminant is $4^{2} - 4 (k) (1) > 0$
$16 - 4 k > 0 ⟹ 4 k < 16$
Therefore, $k < 4$

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| k |

x^{2} + 4 x - k = 0

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