Question

I need help understanding this
Simpliy. Assume all variables are nonnegative

Answer 1

\[\sqrt{x^4y^10}\]

Answer 2

only the 1 and the 0 is acctually ^10

Answer 3

a square root is the same as 1/2 power.

sqrt(x) = x^(1/2)

So for your expression, you have (x^4*y^(10)) all under a square root, so, that whole product is raised to the 1/2 power.

(x^4*y^(10))^(1/2)

When you have a power raised to a power, i.e., (a^n)^m, it is the same as a^(n*m). So for your expression, multiply 1/2 to each power under the radical to simplify the expression.

Answer 4

\(\Large \bf \sqrt{x^4y^{10}}\implies \sqrt[2]{x^4y^{10}}\implies \sqrt[{\color{red}{ 2}}]{(x^2)^{\color{red}{ 2}}y^5)^{\color{red}{ 2}}}\)

what do you think?

Answer 5

woops, missing a (... ahemm \(\Large \bf \sqrt{x^4y^{10}}\implies \sqrt[2]{x^4y^{10}}\implies \sqrt[{\color{red}{ 2}}]{(x^2)^{\color{red}{ 2}}(y^5)^{\color{red}{ 2}}}\)

Answer 6

Wait i don't get it......?

Answer 7

what's confusing you?

Answer 8

\[\sqrt{x^4y^{10}} = (x^4y^{10})^{\frac{1}{2}}\]

Answer 9

Everything... Like how does it change like that?

Answer 10

wait so i time them both by 1/2?

Answer 11

http://www.math-play.com/image-exponents-rules.jpg <---- notice the 3rd rule listed there

Answer 12

It doesn't make sense to me

Answer 13

\[\sqrt{x^4y^{10}} = (x^4y^{10})^{\frac{1}{2}} = x^{4(\frac{1}{2})}y^{10(\frac{1}{2})}\]

Answer 14

Yes, you multiply both powers under the square root by 1/2.

Answer 15

You need to know your exponent rules, as @jdoe0001 pointed out at that site.

Answer 16

so if thats so then i do 4*1/2?

Answer 17

Look at the final expression four posts up. Yes, you multiply both by 1/2. Per exponent rules of raising a power to a power.

Answer 18

oh ok so if you do all that then in the end the answer is \[x ^{2}y^{5}\]

Answer 19

That's right!

Answer 20

can you see if im right on one more question?

Answer 21

Sure.

Answer 22

\[\sqrt{25x^{9}}\]

Answer 23

so the answer for that one would be....... Just a sec

Answer 24

wait i dont get this one....

Answer 25

First, you know you can take 25 out from the radical because you know 5^2 = 25.

Answer 26

So you take out a 5, and it becomes 5*sqrt(x^9)

Answer 27

oh ok then what?

Answer 28

Now, since you know that a square-root is the same as 1/2 power, you do (x^9)^(1/2) = what? You tell me.

Answer 29

\[\sqrt{25x^9} = \sqrt{5*5*x^9} = 5\sqrt{x^9} = 5*(x^9)^{1/2} = ???\]

Answer 30

What did we do last time?

Answer 31

lets see here.... 9 * 1/2 ? do i do that?

Answer 32

Yes, that's what you do.

Answer 33

why does it become \[5\sqrt{x ^{9}}\]

Answer 34

\[\sqrt{25x^9} = \sqrt{5*5*x^9} = 5\sqrt{x^9}\]

Answer 35

What is the square-root of 25?

Answer 36

5

Answer 37

ohhhh i see why now ok

Answer 38

So your answer is?

Answer 39

Remember the rule, sqrt(a) = a^1/2

In this case, your a = x^9

So you need to know the power rule for this. Which we did in your first problem.

You know that (x^n)^m = x^(n*m)

Answer 40

well 1/2*9 is 4.5 im pretty sure..

Answer 41

Right, so put it all together as your final answer.

Answer 42

\[\sqrt{25x^9} = \sqrt{5*5*x^9} = 5\sqrt{x^9} = 5*(x^9)^{1/2} = ???\]

Answer 43

how do you turn 4.5 into a power?

Answer 44

\[\sqrt{25x^9} = \sqrt{5*5*x^9} = 5\sqrt{x^9} = 5*(x^9)^{1/2} = 5x^{9*\frac{1}{2}} = 5x^{\frac{9}{2}} = 5x^{4.5}\]

Answer 45

Trace the steps from the left to the right, arriving at your final, simplified form. Do you understand each step?

Answer 46

my final simplified answer would be\[\sqrt{25x ^{9}}\rightarrow \sqrt{5^{2\left( x ^{4^{}} \right)^{2}}}x \rightarrow5x ^{4\sqrt{x}}\] i Think thats right?

Answer 47

\[\sqrt{25x^9} = \sqrt{5*5*x^9} = 5\sqrt{x^9} = 5*(x^9)^{1/2} = 5x^{9*\frac{1}{2}} = 5x^{\frac{9}{2}} = 5x^{4.5}\]

These are all the steps, and the far-right is your final answer. I'm not sure how I can better explain this so you will understand, and either I am not being clear enough, or you are lacking some fundamental understanding of definitions. Either way, I'm not sure I can be of any further help to you, so perhaps someone else will come along assist you in a different way.

Answer 48

@jdoe0001

Answer 49

ok well thanks

Answer 50

:)

Answer 51

Sorry, and good luck!

Answer 52

you helped me understand it better thx

Answer 53

In the mean time, I suggest you study up on your exponent rules:

http://www.purplemath.com/modules/exponent.htm

Answer 54

ditto

Answer 55

k