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

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Solution

To prove: The equation $x^{2} + p x - 1 = 0$ has real and distinct roots for all real values of $p$ .

Consider $x^{2} + p x - 1 = 0$

Discriminant $D = p^{2} - 4 (1) (- 1) = p^{2} + 4$

We know $p^{2} \geq 0$ for all values of $p$

$\Rightarrow p^{2} + 4 \geq 0$ (since $4 > 0$ )

Therefore $D \geq 0$

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

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