Find the exact value... radicals.

Having a hard time finding the answer to the problem. Can someone help me with this. (6-1)^2 + (15-3)^2 all of this is under a radical sign. I think the answer is 16.46. But I am not sure. I started out by squaring the two sets getting 36+1 +225+9 which equals 271 under a radical sign giving me the answer 16.46. Is this correct

Re: Find the exact value... radicals.

Quote:

Originally Posted by

**modecasec** Having a hard time finding the answer to the problem. Can someone help me with this. (6-1) + 15-3)2 all of this is under a ridical sign. I think the answer is 16.46. But I am not sure. I started out by squaring the two sets getting 36+1 +225+9 which equals 271 under a radical sign giving me the answer 16.46. Is this correct

Hi modecasec,

There's a problem with your notation. Your parentheses aren't balanced. You're missing one left parenthesis. Check you problem and edit your post.

What you have is $\displaystyle \sqrt{(6-1)+15-3)^2$ and that is incorrect.

It could be $\displaystyle \sqrt{((6-1)+(15-3))^2}=17$, or

$\displaystyle \sqrt{(6-1)+(15-3)^2}=\sqrt{149}$. I'm just guessing at what you intended, so edit your post.

Re: Find the exact value... radicals.

yes I messed that up. The two sets in parenthesis are raised to the secodn degree. what do you get now.

Re: Find the exact value... radicals.

Hello, modecasec!

It's hard to read what you typed, but I'll take a guess . . .

Quote:

$\displaystyle \sqrt{(6-1)^2 + (15-3)^2} $

First of all, be careful . . . Don't make up your own rules!

. . $\displaystyle (6-1)^2$ is not equal to $\displaystyle 6^2+1^2$

Look at what we have:

. . $\displaystyle (6-1)^2 \:=\:5^2 \:=\:25$

. . $\displaystyle (15-3)^2 \:=\:12^2 \:=\: 144$

So we have: .$\displaystyle \sqrt{25 + 144} \:=\:\sqrt{169} \:=\:\boxed{13}$ . **

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Another opportunity to make a fatal error . . .

. . $\displaystyle \sqrt{25 + 144}$ .is not equal to .$\displaystyle \sqrt{25} + \sqrt{144}$

Very tempting, isn't it? .Both 25 and 144 are squares.

Well, *of course they* are!

. . They just came from $\displaystyle 5^2$ and $\displaystyle 12^2$, didn't they?

This tempting error appears every time we use the Pythagorean Theorem:

. . $\displaystyle a^2 + b^2 \:=\:c^2$

Think about it.