Find the value of k such that the quadratic equation $5 x^{2} + k x + 4 = 0$ has real and equal roots.

k = \pm \sqrt{5}

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k = - 4 \sqrt{5}

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k = 4 \sqrt{5}

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k = \pm 4 \sqrt{5}

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Solution

The correct option is D $k = \pm 4 \sqrt{5}$
The discriminant of the quadratic equation $a x^{2} + b x + c = 0$ , where a, b, c are real numbers is $D = b^{2} - 4 a c$ .
Thus, the discriminant of the equation $5 x^{2} + k x + 4 = 0 i s k^{2} - 4 (5) (4) = k^{2} - 80$ .

For real and equal roots, D = 0. Hence, $k^{2} - 80 = 0$ .
$\Rightarrow k^{2} = 80$
$\Rightarrow k = \pm 4 \sqrt{5}$

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Let $k$ be the non-zero real number such that the quadratic equation $k x^{2} + 2 x + k = 0$ has two distinct real roots $α and β (α < β) .$

If $α < 2$ and $β > 5$ , then

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