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

  • MHB
  • Thread starter Ziggletooth
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In summary, the previous method failed because it was trying to compare 2 * sqr(16) to 64 which is not equal.
  • #1
Ziggletooth
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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.
 
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  • #2
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.
 
  • #3
Ziggletooth said:
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 \(\displaystyle \sqrt{9(2)}= 3\sqrt{2}\).
 
  • #4
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
 

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

1. What is the formula for solving "Evaluate 3 * sqr(2)"?

The formula for solving "Evaluate 3 * sqr(2)" is to first find the square root of 2, which is approximately 1.414. Then, multiply 3 by this value to get the final answer of approximately 4.242.

2. How do I evaluate the square root of a number?

To evaluate the square root of a number, you can use a calculator or manually find the square root by repeatedly trying different numbers until you find one that, when multiplied by itself, gives the original number. For example, the square root of 16 is 4 because 4*4=16.

3. What does the asterisk symbol mean in "Evaluate 3 * sqr(2)"?

The asterisk symbol (*) is the mathematical symbol for multiplication. In this case, it indicates that we are multiplying 3 by the square root of 2.

4. What is the purpose of "Evaluate" in the phrase "Solving Evaluate 3 * sqr(2): Understanding the Method"?

The word "Evaluate" in this phrase means to find the numerical value or solution of an expression or equation. In this case, we are finding the numerical value of the expression 3 * sqr(2).

5. How can understanding the method of solving "Evaluate 3 * sqr(2)" be useful in real life?

Understanding the method of solving "Evaluate 3 * sqr(2)" can be useful in real life in situations where you need to quickly find the value of an expression involving square roots, such as calculating the length of a diagonal or the area of a circle. It can also be helpful in understanding more complex mathematical concepts and solving more advanced equations.

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