Taking perfect squares out of square roots.
I am not understanding how it works. I don't know if I'm doing it right and the system is gliching or if I'm doing it wrong. More than likely wrong. I need help!!!!
I am not understanding how it works. I don't know if I'm doing it right and the system is gliching or if I'm doing it wrong. More than likely wrong. I need help!!!!
So let's say you have sqrt(20). The first thing to think about are the factors of 20. When thinking about the factors, look for a pair that has a perfect square.
Factors of 20:
1 and 20
2 and 10
4 and 5
Well 4 is a perfect square, so now we can think of sqrt(20) like this:
sqrt(4*5)
We haven't changed the value under the square root sign, we just broke 20 apart. Since the square root of 4 is 2, we can now write the answer as:
2*sqrt(5)
We took the 4 out of the root and made it a 2 since we know what the square root of 4 is.
I hope this helps! Try these two and tell me what you get, or I can explain them if you are still confused:
sqrt(18)
sqrt(72)
Hi Rachel! Thank you for taking the time to explain perfect squares in square roots. I did the two examples you gave me.
Sqrt(18)=3*Sqrt(2)
Sqrt(72)=2*Sqrt(18), 3*Sqrt(8), and 6*Sqrt(2).
I think I'm starting to get perfect squares. I'm seeing the pattern. I also don't understand how they pull out the variables and exponents. I would appreciate any help you would be willing to give. Thanks a bunch, KristinaLeigh
You're welcome! And you got your answers right! I like how you gave multiple ways to factor out squares on the second example. In most cases you will want to simplify completely, meaning you would give the answer of "6*sqrt(2)." (Simplifying completely means you try and get the smallest possible number under the radical.) But it is also useful to break the same number into different squares because you may need different combinations to solve an algebraic equation.
About variables, they are handled in a similar manner.
Let's say we have:
sqrt(y^3)
This could also be written as
sqrt(y*y*y)
because y to the third power is the same as y times itself three times.
So we have a lot of y variables, and we can take some out. The answer would look like:
y*sqrt(y)
This works because the square root of any value to the second power is the value, such as:
sqrt(y^2) is just y
So we can take a y^2 out of the y^3, and we will have one y left under the square root sign.
I hope this helps as well! Here are two examples if you want to do them:
sqrt(y^5)
sqrt(x^3 y^2)
Hi Rachel!!! Thank you again for taking the time to heIp learn this. I figured out how to do it with variables and exponents by studying the problem they gave and seeing how they did. I'm not sure I fully get but I get enough that I have the notes I took to be able to some easy ones. Now it's on to nested fractions. I don't remember doing these in high school. It's very new to me. Thank you, KristinaLeigh
You are very welcome! If you need help with anything else please let me know. Best of luck in further math endeavors!
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