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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.

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

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Be cautious about the negative sign,

$$\frac{3x^{2}}{-\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}}=- \frac{3x^{2}}{\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}} =- \frac{3x^{2}}{3\sqrt{2}} = -\frac{x^{2}}{\sqrt{2}}$$

Maybe, there is a typo in the textbook.

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