Cuberoot, etc.

[elseif]

I have a TI-30XS which is the calculator that I must use when I take the CLEP exam.

And I have no idea how to solve the following problem without foreknowing that I should use "2" (because 2^4 = 16).

See the attached file.

I need to reduce the 48 on the right side to 3.
When I use my calculator and type 4, x sqrt, 48, I get this answer: 2.632148026

How do I get the answer I need?.. I need "3" under the sqrt.

 

[MarkFL]

Hello, and welcome to FMH!

[MATH]48=3\cdot2^4[/MATH]
And so:

[MATH]8\sqrt[4]{3}-\sqrt[4]{48}=8\sqrt[4]{3}-2\sqrt[4]{3}=6\sqrt[4]{3}[/MATH]

 

[JeffM]

I have a TI-30XS which is the calculator that I must use when I take the CLEP exam.

And I have no idea how to solve the following problem without foreknowing that I should use "2" (because 2^4 = 16).

See the attached file.

I need to reduce the 48 on the right side to 3.
When I use my calculator and type 4, x sqrt, 48, I get this answer: 2.632148026

How do I get the answer I need?.. I need "3" under the sqrt.

Simplifying [MATH]8\sqrt[4]{3} - \sqrt[4]{48}[/MATH]
DOES NOT REQUIRE ANY FOREKNOWLEDGE of anything except the fundamental theorem of arithmetic.

[MATH]48 = 2 * 24 = 2^2 * 12 = 2^3 * 6 = 2^4 * 3 \implies \sqrt[4]{48} = \sqrt[4]{2^4 * 3} = 2 \sqrt[4]{3}.[/MATH]
[MATH]\therefore 8\sqrt[4]{3} - \sqrt[4]{48} = 8\sqrt[4]{3} - 2\sqrt[4]{3} = 2\sqrt[4]{3}(4 - 1) = 6 \sqrt[4]{3}.[/MATH]
Of course, if you already happen to know and remember that 48 = 3 * 16 and 16 = 2^4, you can skip over the fundamental theorem of arithmetic. But you do not need that information if you know the fundamental theorem of arithmetic.

 

[elseif]

Hello, and welcome to FMH!

[MATH]48=3\cdot2^4[/MATH]
And so:

[MATH]8\sqrt[4]{3}-\sqrt[4]{48}=8\sqrt[4]{3}-2\sqrt[4]{3}=6\sqrt[4]{3}[/MATH]

Thanks for the welcome.

Your answer requires that I already know 3*2^4 = 48. What if I don't already know that? How do I get to that point, using my calculator?

Another example, which should make my problem very clear:
If I put SQRT 1260 into my calculator, the answer is 35.4964787.
But the answer that I need is 6 sqrt 35.
How do I make my calculator give me the answer 6 sqrt 35?

 

[elseif]

Simplifying [MATH]8\sqrt[4]{3} - \sqrt[4]{48}[/MATH]
DOES NOT REQUIRE ANY FOREKNOWLEDGE of anything except the fundamental theorem of arithmetic.

[MATH]48 = 2 * 24 = 2^2 * 12 = 2^3 * 6 = 2^4 * 3 \implies \sqrt[4]{48} = 2 \sqrt[4]{3}.[/MATH]
[MATH]\therefore 8\sqrt[4]{3} - \sqrt[4]{48} = 8\sqrt[4]{3} - 2\sqrt[4]{3} = 2\sqrt[4]{3}(4 - 1) = 6 \sqrt[4]{3}.[/MATH]

I think I understand that. Take the smallest number that the remainder is divisible by, for as long as I can, right?

See my last post, about sqrt 1260. That one would take a little while to do, using this method. How do I make my calculator give the response that I need?

 

[elseif]

... Someone I know sent me a link to a video that explained how to easily sqrt 1260, by hand. I'm still curious though whether I can make my calculator give the answer in the format that I need.

 

[Steven G]

Unfortunately some calculators can do what you are hoping yours will do, but not knowing your calculator I do not know if it can do as you desire. Maybe look at the manual and see if it does this type of computation in the form you want.

 

[Steven G]

I have a TI-30XS which is the calculator that I must use when I take the CLEP exam.

And I have no idea how to solve the following problem without foreknowing that I should use "2" (because 2^4 = 16).

See the attached file.

I need to reduce the 48 on the right side to 3.
When I use my calculator and type 4, x sqrt, 48, I get this answer: 2.632148026

How do I get the answer I need?.. I need "3" under the sqrt.

If the clep exams have multiple choices then you can use your calculator even with the decimals answers. All you have to do is calculate each given choice and see if your calculator gives the same decimal answer as the answer you got.

 

[elseif]

If the clep exams have multiple choices then you can use your calculator even with the decimals answers. All you have to do is calculate each given choice and see if your calculator gives the same decimal answer as the answer you got.

Nice idea!

 

[lev888]

Your answer requires that I already know 3*2^4 = 48. What if I don't already know that? How do I get to that point, using my calculator?

Your calculator can help you do the work.
You should be able to calculate prime factorizations of numbers. Look it up. It involves dividing your number by primes. If the number is large you can use your calculator.
E.g. 1260.
1260/2=630, 2 is a factor.
630/2=315, 2 is a factor.
315/3=105, 3 is a factor.
105/3=35, 3 is a factor.
35/5=7, 5 and 7 are factors.
So, 1260=2*2*3*3*5*7. 2*2*3*3 is a square.

 

[JeffM]

Your calculator can help you do the work.
You should be able to calculate prime factorizations of numbers. Look it up. It involves dividing your number by primes. If the number is large you can use your calculator.

Elseif

I am sorry that I presumed that you knew the fundamental theorem of arithmetic.

That says that every whole number greater than 1 is either a prime number or is the product of a unique set of prime numbers. (A prime number is a positive whole number that has exactly two distinct divisors, itself and one.)

It is true that factoring large numbers takes a lot of time. It is in fact the basis for some cryptographic systems.

However, in school and test problems, the numbers are usually small and nice enough that you can do them quickly if you apply some basic number facts.

If the final digit of a number is 0 or 5, then five is a factor at least once.

[MATH]1260 \div 5 = 252 \therefore 1260 = 5 * 252.[/MATH] Easy.

The number 252 ends in an even digit other than 0, therefore there are no more factors of 5, but there is at least one factor of 2.

[MATH]252 \div 2 = 126 \text { and } 126 \div 2 = 63 \implies 1260 = 2^2 * 5 * 63.[/MATH]
63 ends in an odd digit so there are no more factors of 2. Is 63 prime or does it have divisors other than itself and 1? Now you may remember that 63 = 7 * 9 = 7 * 3^2. If you remember that, you can complete the factorization immediately. If you do not "see" that, you start dividing by the odd primes other than 5 from smallest on up. So we start with 3.

[MATH]63 \div 3 = 21 \text { and } 21 \div 3 = 7.[/MATH]
So 3 is a factor twice, and 7 is a prime so we are done.

[MATH]1260 = 2^2 * 3^2 * 5 * 7 \implies \sqrt{1260} = 2 * 3 \sqrt{5 * 7} = 6\sqrt{35}.[/MATH]
Once you are used to it, you can do it quickly if the number is "nice."

 

[pka]

Your answer requires that I already know 3*2^4 = 48. What if I don't already know that? How do I get to that point, using my calculator?
Another example, which should make my problem very clear:
If I put SQRT 1260 into my calculator, the answer is 35.4964787.
But the answer that I need is 6 sqrt 35.
How do I make my calculator give me the answer 6 sqrt 35?

Do you understand that some colleges use the CLEP exam as a placement tool?
If you do not already know that \(\displaystyle 1260=2^2\cdot 3^2\cdot 5\cdot 7\) or how to find that on a calculator then you need to be placed accordingly.

 

[Mr. Bland]

[MATH]8\sqrt[4]{3} - \sqrt[4]{48}[/MATH]

I need to reduce the 48 on the right side to 3.
When I use my calculator and type 4, x sqrt, 48, I get this answer: 2.632148026
How do I get the answer I need?.. I need "3" under the sqrt.

@lev888 has the right idea: you can reduce figures inside a radical by finding the prime factors of that figure.

In the case of [MATH]\sqrt[4]{48}[/MATH], the prime factors of 48 are 2 * 2 * 2 * 2 * 3. Since 2 appears four times (the power of the root you're trying to find), you can "extract" it from the radical to get [MATH]2\sqrt[4]{3}[/MATH].