Show that the equation $x^{2} + a x + 1 = 0$ has real and distinct roots for all real values of $a$ .

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Solution

$D = b^{2} - 4 a c$

$D = a^{2} - 4 (1) (1) = a^{2} - 4$

for all the real values of $a > 2 a n d a < - 2$

$D > 0$

Hence $x^{2} + a x + 1 = 0$ has real and distinct roots for all the real values of a

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