Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0/Zero Discriminant
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Theorem
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c = 0$.
Then:
- $\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
is not defined.
Proof
Suppose that $b^2 - 4 a c = 0$.
Then:
\(\ds a x^2 + b x + c\) | \(=\) | \(\ds \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b}^2} {4 a}\) | as $b^2 - 4 a c = 0$ |
But we have that:
- $\paren {2 a x + b}^2 > 0$
while under our assertion that $a < 0$:
- $4 a < 0$
and so:
- $a x^2 + b x + c < 0$
Thus on the real numbers $\sqrt {a x^2 + b x + c}$ is not defined.
Hence it follows that:
- $\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
is not defined.
$\blacksquare$