Show that the equation $x^{2} + a x - 4 = 0$ has real and distinct roots for all real values of $a$ .

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Solution

The given equation is $x^{2} + a x - 4 = 0$

Comparing given equation with $a x^{2} + b x + c = 0$

$∴ a = 1, b = a, c = - 4$

The discriminant of the given equation is given by $D = b^{2} - 4 a c = a^{2} - 4 \times - 4 = a^{2} + 16$

Clearly, $D = a^{2} + 16 > 0$ for all $a \in R$

Hence, the given equation has real and distinct roots.

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