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I have been given the following equation, semi-derived from the quadratic equation:

$\frac{+\sqrt{b^2-a}}{a}<\frac{-\sqrt{b^2-a}}{a}$

I need to prove that ${a}<0$ is a possible real solution to this equation. Wolfram Alpha has verified that this is true, but I am not sure how to derive this.

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  • $\begingroup$ Riley: a <0 , the 2 roots are defined (why).Now:LHS:divide a positive number by a, a negative number , result negative, hence LHS negative ,RHS positive(why?), inequality is fine. $\endgroup$ – Peter Szilas Mar 19 at 15:10
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If $a$ is positive, the left side is a positive number (pos/pos = pos) and the right side is a negative number (neg/pos = neg), so the inequality can never be satisfied (pos < neg is never true).

You can also eliminate the possibility that $a=0$ since neither side is then defined (division by zero would occur on both sides, which is not permitted).

The remaining possibility is that $a$ is negative.

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One fraction is just the negative of the other. So (assuming $b^2 - a>0$) it becomes a matter of figuring out which is positive and which is negative. Since square roots by definition are positive, we get ...

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First, note that for real values of $b$ that $b^2 \geq 0$ thus any $a < 0$ will make the $b^2-a$ term a positive number. The rest follows by considering the denominator and the sign change in the numerator.

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