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Manhattan GMAT Quant (Algebra Book - Chapter 4 Problem Set)

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Manhattan GMAT Quant (Algebra Book - Chapter 4 Problem Set)

by stevennu » Mon Aug 19, 2013 11:15 pm
Question 11

Simplify: sqrt(x^2y^3 + 3x^2y^3)
The answer is 2xy sqrt (y)

I'm wondering why the expression inside the square root is not simplified to (4x^2)(2y^3). Why is it necessary to view the entire expression as two separate parts on either side of the + sign? The book simplifies the expression inside the square root to 4x^2y^3. I just don't understand why we don't assume that there are "1's" in front of the Y and then add them together in order to get 2y^3.

Can someone please help?
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by sanju09 » Tue Aug 20, 2013 1:50 am
stevennu wrote:Question 11

Simplify: sqrt(x^2y^3 + 3x^2y^3)
The answer is 2xy sqrt (y)

I'm wondering why the expression inside the square root is not simplified to (4x^2)(2y^3). Why is it necessary to view the entire expression as two separate parts on either side of the + sign? The book simplifies the expression inside the square root to 4x^2y^3. I just don't understand why we don't assume that there are "1's" in front of the Y and then add them together in order to get 2y^3.

Can someone please help?
x^2y^3 is a single term, let's call it equal to z.

Now sqrt (z + 3z) = sqrt (4z) = 2 sqrt z

= 2 sqrt (x^2y^3) = 2 sqrt (x^2y^2.y)

= 2 xy sqrt (y).

It'll be nice idea if you do little work on operations on algebraic expression or simply give x, y some values and test answers in both situations, to understand the rules better. The given expression is not calling to add y^3 terms separately, x^2 is also there as factor of the whole term; so we cannot do that as suggested by you.
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by ganeshrkamath » Tue Aug 20, 2013 3:07 am
stevennu wrote:Question 11

Simplify: sqrt(x^2y^3 + 3x^2y^3)
The answer is 2xy sqrt (y)

I'm wondering why the expression inside the square root is not simplified to (4x^2)(2y^3). Why is it necessary to view the entire expression as two separate parts on either side of the + sign? The book simplifies the expression inside the square root to 4x^2y^3. I just don't understand why we don't assume that there are "1's" in front of the Y and then add them together in order to get 2y^3.

Can someone please help?
Ok, let's make the problem simpler.
Let a = x^2, b = y^3
Now you are saying 3ab + ab = (3a+a)(b+b)
Well, this is not right.
The rule is: When you have a common variable in all the terms, you can take it out.
So 3ab + ab = (3a + a) * b
or 3ab + ab = (3b + b) * a
or 3ab + ab = (3 + 1) * ab
All of the above are the same.

Let's take an example:
a = 2 and b = 3
Now 3ab + ab = 3(2)(3) + (2)(3) = 18 + 6 = 24
(3a + a) * b = (6 + 2) * 3 = 8*3 = 24
(3b + b) * a = (9 + 3) * 2 = 12*2 = 24
(3 + 1) * ab = 4 * 2 * 3 = 4*6 = 24
As you can see, all of the above are the same.

But (3a+a)(b+b) = (6 + 2)(3 + 3) = 8*6 = 48 which is not right.

Hope this helps.
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by stevennu » Tue Aug 20, 2013 8:47 am
How do we know that 3x^2y^3 and x^2y^3 should be viewed as two separate terms? Why can't we view them as (3x^2)(1y^3) + (1x^2)(1y^3)? Is it because (3x^2 + 1x^2) (1y^3 + 1y^3) would violate order of operations since we are adding terms before multiplying? I don't know why I'm struggling so much with this concept. I just need to know the rule and reasoning behind it so that I know how to proceed on similar questions.

Thanks for the help and replies thus far.
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by ganeshrkamath » Tue Aug 20, 2013 7:59 pm
stevennu wrote:How do we know that 3x^2y^3 and x^2y^3 should be viewed as two separate terms? Why can't we view them as (3x^2)(1y^3) + (1x^2)(1y^3)? Is it because (3x^2 + 1x^2) (1y^3 + 1y^3) would violate order of operations since we are adding terms before multiplying? I don't know why I'm struggling so much with this concept. I just need to know the rule and reasoning behind it so that I know how to proceed on similar questions.

Thanks for the help and replies thus far.
Ok, I think it's called the distributive rule of multiplication.
It says that
A(B+C) = AB + AC

What you should be doing is the opposite!
i.e. AB + AC = A(B+C)

So (3x^2)(1y^3) + (1x^2)(1y^3) = (1y^3)(3x^2 + 1x^2)
(Here 3x^2 = B, 1x^2 = C, 1y^3 = A)

Hope this helps.
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by stevennu » Tue Aug 20, 2013 9:24 pm
ganeshrkamath wrote:
stevennu wrote:How do we know that 3x^2y^3 and x^2y^3 should be viewed as two separate terms? Why can't we view them as (3x^2)(1y^3) + (1x^2)(1y^3)? Is it because (3x^2 + 1x^2) (1y^3 + 1y^3) would violate order of operations since we are adding terms before multiplying? I don't know why I'm struggling so much with this concept. I just need to know the rule and reasoning behind it so that I know how to proceed on similar questions.

Thanks for the help and replies thus far.
Ok, I think it's called the distributive rule of multiplication.
It says that
A(B+C) = AB + AC

What you should be doing is the opposite!
i.e. AB + AC = A(B+C)

So (3x^2)(1y^3) + (1x^2)(1y^3) = (1y^3)(3x^2 + 1x^2)
(Here 3x^2 = B, 1x^2 = C, 1y^3 = A)

Hope this helps.
Cheers
I actually figured it out earlier. You are basically pulling out the common term and then simplifying. Don't know why I had that road block. That was a concept I learned in high school algebra but apparently lost track of. Thanks
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