Solving Evaluate 3 * sqr(2): Understanding the Method

[Ziggletooth]

I'm having trouble understanding how this method works and why it appears not to work on similar questions.

For the question evaluate 3 * sqr(2)

I understand I can square both factors to eliminate the square root.

(3 * 3) * 2 = 18

However this does not appear to work with 2 * sqr(16)

(2 * 2 ) * 16 = 64

But the answer is 2 * 4 = 8. I can see why that is, because 4 is the sqr(16), but I don't understand why the previous method failed on what appears to me to be essentially the same question.

Any help would be appreciated, thank you.

 

[mrtwhs]

You correctly stated that \(\displaystyle 2 \sqrt{16} \neq 64\).

Why would you believe that \(\displaystyle 3 \sqrt{2}=18\)? These two are not equal in the same way that the first two are not equal.

 

[HOI]

I'm having trouble understanding how this method works and why it appears not to work on similar questions.

For the question evaluate 3 * sqr(2)

I understand I can square both factors to eliminate the square root.

(3 * 3) * 2 = 18

Yes, you can but how does that help you evaluate \(\displaystyle 3\sqrt{2}\)?

I suspect that you are thinking, instead, of "taking the '3' inside the square root":
\(\displaystyle 3\sqrt{2}= \sqrt{9(2)}= \sqrt{18}\).

However this does not appear to work with 2 * sqr(16)

(2 * 2 ) * 16 = 64

But it does work: \(\displaystyle 2\sqrt{16}= \sqrt{4(16)}=\sqrt{64}\)

But the answer is 2 * 4 = 8. I can see why that is, because 4 is the sqr(16), but I don't understand why the previous method failed on what appears to me to be essentially the same question.

Any help would be appreciated, thank you.

You can't just "throw away" the square root. If the problem is to find [tex]a\sqrt{b}[/tex] then you can "take the a inside the square root" to get \(\displaystyle \sqrt{a^2b}\) but you still have to take the square root!

Actually most people would consider "simplifying" a square root to be going the other way: to simplify \(\displaystyle \sqrt{18}\) write it as \(\displaystyle \sqrt{9(2)}= 3\sqrt{2}\).

 

[Ziggletooth]

I'm sorry it appears the source of my confusion was very much that I assumed I was evaluating something when in fact I was originally doing these operations to make it easier to compare which of two expressions were greater. That wasn't clear to me at the time so I was confused why it wasn't working is some cases but it depended on the context of the numbers I was comparing.

Thanks