# Is this a mistake on my part or theirs?

I'm not sure if I'm the one making the mistake, or my math book. It looks like the negative sign completely disappeared.

$$\frac{3x^2}{-\sqrt{18}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3x^2\sqrt{2}}{-\sqrt{36}} = \frac{3x^2\sqrt{2}}{6} = \frac{x^2\sqrt{2}}{2}$$

Here is the original image.

Also, I am new to Stackexchange, so tell me if I am doing something wrong.

• You don't need any points to ask a question on meta. Try using the code, someone will fix it up for you. – Gerry Myerson Jan 28 '16 at 11:53
• Oh... It said I needed 5 reputation to ask a question there, perhaps I was mistaken? But thanks for that bit of info anyhow. – L7vanmatre Jan 28 '16 at 11:54
• But, you're right, it does look as though the minus sign has gone walkabout. – Gerry Myerson Jan 28 '16 at 11:54
• Maybe they slipped a change past me when I wasn't looking, but I always thought anyone could post questions, points or no. – Gerry Myerson Jan 28 '16 at 11:55
• In this case it's actually not such a bad thing to post a picture of the equations, since it is relevant to see the primary source. – Arthur Jan 28 '16 at 12:15

It's probably them. It starts out correct, but when they go to take the negative square root of 36, there was an error. The correct approach is (or should I say, should be): $$\frac{3x^2}{-\sqrt{18}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3x^2\sqrt{2}}{-\sqrt{36}} = \frac{3x^2\sqrt{2}}{-6} = -\frac{x^2\sqrt{2}}{2}$$ This is because $-\sqrt{36}$ is actually $-(\sqrt{36})$. Or you could just apply the fraction rule and make it negative and go from there. It'll still give you the same answer.
$$\frac{3x^{2}}{-\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}}=- \frac{3x^{2}}{\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}} =- \frac{3x^{2}}{3\sqrt{2}} = -\frac{x^{2}}{\sqrt{2}}$$