5
$\begingroup$

I was trying to solve this square root problem, but I seem not to understand some basics.

Here is the problem.

$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2$$

The solution is as follows:

$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2 = \Bigg(\frac{3}{2} - \sqrt{2} - 1 + \sqrt{2}\Bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}$$

Now, what I don't understand is how the left part of the problem becomes: $$\frac{3}{2} - \sqrt{2}$$

Because I thought that $$\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2}$$ equals to $$\bigg(\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2\bigg)^{\frac{1}{2}}$$ Which becomes $$\sqrt{2} - \frac{3}{2}$$

But as you can see I'm wrong.

I think that there is a step involving absolute value that I oversee/don't understand. So could you please explain by which property or rule of square root is this problem solved?

Thanks in advance

$\endgroup$
2
$\begingroup$

Nicely put question.

You are right about the absolute value missing somewhere. Indeed, we have:

$$\sqrt{x^2} = |x|.$$

In your case, we have

$$\sqrt{\left(\sqrt{2}-\frac{3}{2}\right)^2}=\left|\sqrt{2}-\frac{3}{2}\right|.$$

But $\sqrt{2}-\frac{3}{2}$ is negative, so the absolute value "chooses" the positive version of this, that is,

$$\left|\sqrt{2}-\frac{3}{2}\right| = -\left(\sqrt{2}-\frac{3}{2}\right)=\frac{3}{2}-\sqrt{2}.$$

I hope this helps.

$\endgroup$
2
$\begingroup$

This is a mistake that might have been avoided by being aware of the numerical values of some of the expressions involved. A very rough approximation for $\sqrt{2}$ is $1.41$. Clearly $\frac{3}{2} = 1.5$. Then $\sqrt{2} - \frac{3}{2} \approx -0.09$, and that squares to $0.0081$, and the square root of that is $0.09$, not $-0.09$.

At no point in this problem is it actually necessary to take the square root of a negative number. But the possibility of going down that path, even if erroneously, would perhaps have alerted you that some things in this problem should be evaluated rather than canceled. Thus, $$\left(\sqrt{2} - \frac{3}{2}\right)^2 = \left(-\frac{3}{2} + \sqrt{2}\right)^2 = \frac{17}{4} - 3 \sqrt{2} \neq -\frac{3}{2} + \sqrt{2}.$$

I'm not suggesting that you always run through problems with crude numerical approximations. But it can be a good way to check you're on the right track.

$\endgroup$
1
$\begingroup$

A square root is always a non negative number so $\sqrt{x^2}=|x|$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.