Multiplying Monomials with Radical Indices in Algebra II/Trig Honors

[lj19]

I'm in Algebra II/Trig Honors.
How do you multiply monomials with an index in the radical?
Thank you.

 

[HallsofIvy]

Please give an example. What do you mean by "an index in the radical"?

 

[lj19]

An example from my notes is:

(4(sqrt of x to the fifth power)) with a 3 in the index of the radical, multiplied by (sqrt of 16 x squared) with a 3 in the index of the radical. This problem was simplified to 4(16x to the seventh) with a 3 in the index of the radical, and simplified to the answer: 4(8x to the sixth 2x) with a 3 in the index of the radical.

An index is a number on the outside of the radical. If it is not given in the problem, it is automatically 2, but if it is given, then that given number is used.

 

[Inferior89]

Do you mean for example \sqrt[3]{27} ?

 

[lj19]

Yes that is an index of 3 in the radical.

 

[Inferior89]

Is this your problem to simplify? Hard to understand without math notation. No problem just tell me if it is correct.

4 \sqrt[3] {x^5} \cdot \sqrt[3]{16x^2}

 

[lj19]

Yes it is. In my notes I have that problem solved. But I don't understand how to solve it. Could you explain how to? And how do you multiply monomials with an index in the radical?

 

[Inferior89]

Well it is no difference than if you have a normal square root.

For example:

\sqrt[a]{b} \cdot \sqrt[a]{c} = \sqrt[a] {bc}

 

[lj19]

So would the problem turn into:
5(16x to the seventh) and the radical staying as 3?
I'm not using my notes for that.
But I know that there is another step after this to get the final answer.

 

[Inferior89]

For your example

4 \sqrt[3] {x^5} \cdot \sqrt[3]{16x^2} = 4 \sqrt[3]{16 x^7} = 4 \sqrt[3]{2x \cdot 8x^6} =

4 \sqrt[3]{8x^6}\cdot\sqrt[3]{2x} = 4 \sqrt[3]{x^6}\cdot\sqrt[3]{8}\cdot\sqrt[3]{2x} = 4 \cdot x^2 \cdot 2 \cdot \sqrt[3]{2x} = 8x^2 \sqrt[3]{2x}

 

[lj19]

I have questions about this problem. Why doesn't the 1x in the first part and the 16x in the first part not add to 17x when they are multiplied, is it just because the only thing that will add are the exponents and everything else is multiplied [except for the index which stays the same]? Also, when multiplying monomials with an index in the radical, does the index number always stay the same, like it did in this problem?
In my notes, the problem was solved the way it is written above, but the final answer is: 4(8x to the sixth times 2x) with the 3 in the index of the radical, and that is the farthest it is solved.

 

[Inferior89]

1x times 16x is 16x^2.
1x plus 16x is 17x.

Yes the index stays the same.

They didn't simplify it fully in your example then for some reason.. The whole point in writing 16x^7 = 2x times 8x^6 is so you can use that \sqrt[3]{8x^6} = 2x^2.

 

[lj19]

I think for this class, I only have to simplify in it those 2 specific steps.
Could you explain what the index in the radical is? I don't understand it. In class, my teacher explained that the other exponents correspond with the index number to add up to another exponent.

 

[Inferior89]

You know when you have \sqrt{x} = y this means that y \cdot y = x

With the index 3 you get \sqrt[3]{x} = y \Rightarrow y \cdot y \cdot y = x.

This is why \sqrt[3]{8x^6} = 2x^2. Because 2x^2 \cdot 2x^2 \cdot 2x^2 = 8x^6

 

[lj19]

If I put numbers into those letters, I can understand it, I think I just have to understand that it works with letters too. For example, in your first example, sqrt4=2 would mean that 2(2)=4. If I use those numbers in your second example with the index of the 3 in the radical, then sqrt4 [with the 3 in the index]=2=2(2)(2)=4.
Could you fill in numbers for the second example, so I can try to understand it. Since my numbers from the first example didn't work in the second. How is y=y(y)(y)?
Thank you.

 

[Inferior89]

\sqrt[3]{27} = 3 because 3 \cdot 3 \cdot 3 = 27

\sqrt[3]{4} is NOT 2 because 2 \cdot 2 \cdot 2 is not 4.

\sqrt[3]{8} = 2 because 2 \cdot 2 \cdot 2 = 8

\sqrt[3]{x^3} = x because x \cdot x \cdot x = x^3

\sqrt[3]{x^6} = x^2 because x^2 \cdot x^2 \cdot x^2 = x^6.

You can also think as taking the cube root (index 3) is the same as taking the exponent to a third.

So \sqrt[3]{x^6} = (x^6)^{1/3} = x^{6 \cdot 1/3} = x^2

 

[lj19]

Does the index of the radical determine how many exponential variables will be multiplied to get the number in the radical? For example, sqrt of x to the sixth with a 3 as the index in the radical. The three in the index of the radical determines the (xsquared)(xsquared)(xsquared) which equals x to the sixth, which is the number in the radical. Is this correct?

 

[Inferior89]

I don't really know what more to say. Look at my examples I did before and see if you can figure these out:

\sqrt[3]{64}

\sqrt[3]{x^9}

\sqrt[4]{16}

 

[lj19]

I'm not asking for you to explain any more about multiplying monomials with an index in the radical, I understand how to do it. I just wanted to know if in your example, that the 3xto the sixth [with the index of 3] is in that form to be solved, because the index of 3 determines that you need 3 sets of xsquared's which will equal x to the sixth, also being the number in the radical. I noticed that it all corresponds, which may be what my teacher had explained.

1. I got 4, because if you take 4 and cube it you get 64.
2.I got x to the third times x to the third times x to the third which would equal x to the ninth, and the 3 in the index represents the 3 sets of x to the third, which is my final answer.
3. I got 4, because if you split 16 up four ways, then it would be 4.

 

[Doubell]

u guys do calculus so this easy question should be a breeze can u explain to me how u write for example 5x^2 +2x-7 in the form a(x+b)^2 +c

 

[lj19]

I'm sorry, I do not do calculus, but that question doesn't pertain to my thread.

 

[Inferior89]

I'm not asking for you to explain any more about multiplying monomials with an index in the radical, I understand how to do it. I just wanted to know if in your example, that the 3xto the sixth [with the index of 3] is in that form to be solved, because the index of 3 determines that you need 3 sets of xsquared's which will equal x to the sixth, also being the number in the radical. I noticed that it all corresponds, which may be what my teacher had explained.

1. I got 4, because if you take 4 and cube it you get 64.
2.I got x to the third times x to the third times x to the third which would equal x to the ninth, and the 3 in the index represents the 3 sets of x to the third, which is my final answer.
3. I got 4, because if you split 16 up four ways, then it would be 4.

1 and 2 are correct. 3 is wrong because 4 \cdot 4 \cdot 4 \cdot 4 is not 16.

It is a bit hard to understand what you want.. Try rewording what you want to know a bit clearer.

u guys do calculus so this easy question should be a breeze can u explain to me how u write for example 5x^2 +2x-7 in the form a(x+b)^2

Make a separate thread in the correct forum category, please.

 

[lj19]

Sorry if my explanation was confusing.
Thanks for confirming that 1 and 2 were correct, and for correcting 3.
An example you explained earlier in this thread was: sqrt of 3xtothesixth with a 3 as the index in the radical. I mentioned earlier that my teacher said that the index corresponds with the exponents in the problem, which I realized they did in that problem. You simplified the problem to xsquared times xsquared times xsquared and then explained that, that would equal xtothesixth.
I noticed what my teacher tried to explain I think, with the numbers corresponding, because the index of 3 created 3 "xsquares" and those 3 "xsquares" equal x6, which is the number originally in the radical.
That's the best I can explain it. I just mean how the index [3 in that case], being 3, 4 or, 5, will determine how many x to an exponent you multiply with [x squared multiplied 3 times in that case], and then the end result of that, is the number from the radical [x6 in that case].
Thank you for helping me.

 

[Inferior89]

I think what your teacher means with that 'the index corresponds to the exponent' is that

\sqrt[a]{b} = b^{1/a}
You see that the index a corresponds to the a in the exponent.

So for example \sqrt[3]{x^6} = (x^6)^{1/3} = x^{6 \cdot 1/3} = x^2

 

[lj19]

I think that's an example of what I was explaining, how in your example the 3 in the index could relate to the step: xsquared*xsquared*xsquared. Is xsqaured the final step in the problem, and is that the answer?
In an earlier example of the problem you wrote xsquared*xsquared*xsquared=x6? Is that to check that that number equals the one in the radical?
And in your newest example, I think for class we solve it differently, but get the same answer.

So I think my questions are:
Does the index in the radical determine how many times the problem is broken up, like how the index was 3, and the problem simplified to xsquared*xsquared*xsquared?
And is xsquared the final answer? And is the x to the sixth after the xsquared*xsquared*xsquared just to check that it matches the number in the radical?

 

[Inferior89]

The index tells me that how many times I need to multiply something to get the value inside the radical.

You see I multiply x^2 three times because the index was 3. If the index was four I would have to multiply SOMETHING ELSE, four times to get what was inside the radical.

So if it says \sqrt[5]{54} you need to ask yourself 'what number multiplied with itself 5 times will be 54. In other words what number to the power of 5 will be 54. That will be the result of \sqrt[5]{54}. For this example it is not an integer and likely not even a rational number.

 

[lj19]

I understand what you're explaining, that's what I was trying to explain, how the number in the radical determines "how many times I need to multiply something to get the value inside the radical".
So if the number was the sqrt of x to second power with a 4 as the index, you would have to multiply (x)(x)(x)(x). And to check that you would get x*x*x*x=x to the fourth. Is that example correct the way I solved it?
Or would the way I solved that only work if it was x to the fourth with a 2 as the index?
Thank you.

 

[Inferior89]

\sqrt[4]{x} would not be x since x \cdot x \cdot x \cdot x is not x.

You need to ask yourself, what do I need to take to the power of 4 to get x. This will be the answer to what \sqrt[4]{x} is.

Since this can be hard you need to learn that \sqrt[4]{x} = x^{1/4}

 

[lj19]

How would you sovle x squared with a four as the index?

 

[Inferior89]

\sqrt[4]{x^2} = (x^2)^{1/4} = x^{2 \cdot 1/4} = x^{1/2} = \sqrt{x}, \quad x \geq 0

 

[lj19]

Thanks, I think that explanation has to do with another part of the lesson in my notes called "Finding the smallest value and smallest integral value"?
What monomial would give me x*x*x*x then? I know it would have to have an index of four. But what exponent would it have to have?

 

[Inferior89]

x*x*x*x = x^4

\sqrt[4]{x^4} = x

\sqrt[4]{x^4} = (x^4)^{1/4} = x^{4\cdot 1/4} = x^1 = x