Adding and Subtracting Radicals + Rationalizing the Denominator Confusion

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GetThroughDiffEq

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Okay, I've literally done these problems over three times and reread the chapter.

I'm assuming there has to be something wrong with how I divide the terms.

Here are the problems which have eluded me (not homework problems):

31. \[\frac{√200^3}{√10x^-1}\]

My answer: 2x√5

Correct Answer: 2x^2√5

My thinking: I ignored the ^-1 on 10x because -10x*-10x (100x^2) is positive.

First I simplified, to (√20x^2).

Second time around to √(2)(2)(5)(x)(x) -- crossed out one of each 2 and x to get 2x√5
39. \[√50x-√8x\]

My answer: 3√7x

Correct Answer: 3√2x

My thinking: Subtracted to get √42x. Simplified to √(3)(3)(7)(x). Therefore, 3√7x
 
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43.
\[3√8-√32+3√72-√75\]

My answer: (10√8)-(5√3)

Correct Answer: (20√2)-(5√3)

My thinking: 1. Left 3√8 alone.
2. Simplified √32 to √(8)(2)(2). Thus, 2√8.
3. Simplified 3√72 to 3√(8)(3)(3). Thus, 9√8.
4. Simplified √75 to √(5)(5)(5)(3). Thus, 5√15 (Before, I got 5√3, somehow).
5. Therefore, (10√8)-(5√3) (or (10√8)-(5√15), in this example).

Rationalizing the Denominator:

51. \[\frac{7}{(√5)-(2)}\]

My Answer: (√21)/(3)

Correct Answer: 7(√5+2)

53. \[\frac{6}{(√5)+(√3)}\]

My Answer: (3√2√30)/15)

Correct Answer: 3(√5-√3)

Thanks in advance.
 
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1) The notation is very confusing. Please use parentheses to clarify. Use TOO MANY. Or learn just a little LaTeX.

2) 10x^-1 probably has nothing to do with positive or negative. Guessing: 10x^-1 = [math]10x^{-1} = \frac{10}{x}[/math]
 
tkhunny said:
1) The notation is very confusing. Please use parentheses to clarify. Use TOO MANY. Or learn just a little LaTeX.

2) 10x^-1 probably has nothing to do with positive or negative. Guessing: 10x^-1 = [math]10x^{-1} = \frac{10}{x}[/math]
Click to expand...

I will do that from now on; thanks for the suggestion. Unfortunately, Windows Defender just blocked one of the installations, so I'll need to research more into it.

Note: Fixed the notation. Apparently you don't allow multiple numbers to be * together with a sqrt i.e. 5x8x2x2 (replace x with *).
 
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GetThroughDiffEq said:
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Okay, I've literally done these problems over three times and reread the chapter.

I'm assuming there has to be something wrong with how I divide the terms.

Here are the problems which have eluded me (not homework problems):

31. (√200^3)/(√10x^-1)

My answer: 2x√5

Correct Answer: 2x^2√5

My thinking: I ignored the ^-1 on 10x because -10x*-10x is positive. First I simplified, to √20x^2.

Second time around to √225xx -- crossed out one of each 2 and x to get 2x^5
____
39. (√50x)-(√8x)

My answer: 3√7x

Correct Answer: 3√2x

My thinking: Subtracted to get √42x. Simplified to √337*x. Therefore, 3√7x
Click to expand...
1st of all the ^-1 is NOT on the 10x using your language
Even if it was, why do you think (10x)^-1 = (-10x)^2 which equals what you said, namely (-10x)(-10x)??????

a^-1 = 1/a^1 = 1/a

10x^-1 = 10/x NOT (-10x)(-10x)

(√200^3)/(√10x^-1) = √(200^3x)/(√10)= 200√(200x)/(√10)= 200√(20x) =200√(45x) =400√(5x)
 
√(50x) - √(8x) = √(25*2x) - √(4*2x) = 5√(2x) - 2√(2x) = (5 - 2)*√(2x) = 3√(2x)

EDIT: Inserted * symbols
 
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GetThroughDiffEq said:
31. [MATH]\frac{√200^3}{√10x^-1}[/MATH]
My answer: 2x√5

Correct Answer: 2x^2√5

My thinking: I ignored the ^-1 on 10x because -10x*-10x (100x^2) is positive.

First I simplified, to (√20x^2).

Second time around to √(2)(2)(5)(x)(x) -- crossed out one of each 2 and x to get 2x√5
39. [MATH]√50x-√8x[/MATH]

My answer: 3√7x

Correct Answer: 3√2x

My thinking: Subtracted to get √42x. Simplified to √(3)(3)(7)(x). Therefore, 3√7x
Click to expand...

I tried to fix up your attempt at using Latex, but there's more wrong than that, so I left it mostly as it was. You omitted an "x" in the first problem, which should have been [MATH]\frac{\sqrt{200x^3}}{\sqrt{10x^{-1}}}[/MATH]. The second should be [MATH]\sqrt{50x}-\sqrt{8x}[/MATH].

But the main error on the first is that you can't ignore a negative exponent. Just do with it what you've been taught. (There are a couple ways to carry out the details, such as moving a negative power to the numerator and changing the sign.)

For the second, you have to simplify each radical and then factor out the common radical. You can't just subtract the radicands!
 
GetThroughDiffEq said:
… I've … reread the chapter …

39. √(50x) - √(8x)

… My thinking: Subtracted to get √42x. Simplified to √(3)(3)(7)(x). Therefore, 3√7x
Click to expand...
Hello. Please don't invoke LaTeX delimiters, if you're not going to use LaTeX. Also, note the grouping symbols inserted in red, above. These are important to type, whenever the radicand contains more than one number. In other words, those grouping symbols are needed to show readers that x appears in the radicand versus following the radical.

We cannot subtract (or add) numbers in separate radicands, like you did (i.e., you can't subtract 8 from 50.) Check your textbook's index for properties of radicals. There ought to be a listing of basic operations to show what you're allowed to do when simplifying. Print it out, if you can, for reference as you work. If you try doing stuff that's not on that list, then you're probably going off the rails. ;)

Also, how did you get 42 = 3∙3∙7 ?

Have you memorized the multiplication table, yet? These types of exercises take much longer for students who can't recognize products which appear in the multiplication table.

Are you familiar with prime factorizations (and how to find them)? For examples:

50 = 2∙52

8 = 23

?

EDIT: Added suggestion to look up listing of radical properties, for reference.
 
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Otis said:
Have you memorized the multiplication table, yet? These types of exercises take much longer, when you can't recognize products which appear in the multiplication table.
Click to expand...
:giggle:?:):D:geek:??
 
Jomo said:
(√50x)-(√8x) =√(25*2x)-√(42x) = 5√(2x)-2√(2x) = (5-2)(√2x) = 3√(2x)
Click to expand...

Why are you subtracting √(42x)? It makes no sense to me at all.

I tried these four problems again and I keep getting them wrong. I guess it's time to move on:

For the rationalizing the denominator questions, I'm still stuck (although, I've gotten the basic ones right).

On Question #51, for example, I tried multiplying both the numerator and denominator with:

1. √5 alone AND 2. (√5)-(2)

Please tell me, for the rationalizing the denominator, what am I doing WRONG?
 
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For #51, you multiply numerator and denominator by √5 + 2, the conjugate.
 
GetThroughDiffEq said:
Why are you subtracting √(42x)? It makes no sense to me at all …
Click to expand...
There's a typo, in that post. Jomo meant to type √(4*2x) because 4×2=8.

EDIT: Maybe Jomo had typed it correctly; I'm trying to fix the post, but a system error keeps removing my * symbols … I've inserted the correction via LaTeX.

?
 
GetThroughDiffEq said:
Why are you subtracting √(42x)? It makes no sense to me at all.

I tried these four problems again and I keep getting them wrong. I guess it's time to move on:

For the rationalizing the denominator questions, I'm still stuck (although, I've gotten the basic ones right).

On Question #51, for example, I tried multiplying both the numerator and denominator with:

1. √5 alone AND 2. (√5)-(2)

Please tell me, for the rationalizing the denominator, what am I doing WRONG?
Click to expand...
It should have been \(\displaystyle \sqrt{4*2x}\) not \(\displaystyle \sqrt{42x}\) sorry about that.
 
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