# Why is a < 0 the only solution to the following inequality?

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.

• 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. – Peter Szilas Mar 19 at 15:10

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.
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 ...
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.