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Prove that the quadratic equation

p x^{2} + (2 p + 1) x + p + 1 = 0

has exactly two solutions for all real values of $p$ .

Hint

Express the discriminant in terms of $p$ and use completing the square to find its minimum value.

Solution

The discriminant of

p x^{2} + (2 p + 1) x + p + 1 = 0

\begin{aligned} Δ & = (2 p + 1)^{2} - 4 p (p + 1) \\ = 4 p^{2} + 4 p + 1 - 4 p^{2} - 4 p \\ = 1 \end{aligned}

We see that, regardless of the value of $p$ , $Δ = 1 > 0$ . Since the discriminant is therefore positive, the quadratic has two disctinct roots for all $p \in R$ .