## Math Problem: who is incorrect, me or the book?

• Posted by a hidden member.
[EDIT] Problem solved thanks to help from others. In the end, the book didn't include and absolute sign in Column A, which would make the book correct if it did.

This book and I disagree on an answer to a question (intended for practice for the GRE).

Here is the question (FYI for the four answers are referred to from top to bottom as answer A, B, C, & D.):

So I plugged in numbers and here's what I found.

If x = 4, y = 9, then Column B is greater than Column A (My math showed that Column A = -1 and Column B = 1).
However, if x = 9, y = 4, Column A is equal to Column B, or the two quantities are equal (both equal 1)

So my conclusion was this: depending upon the number, Column B is either equal to or greater than Column A. So neither answers A, B, or C are correct, leaving me to conclude that only Answer D is the correct answer (The relationship cannot be determined from the information given).

But the book says the answer is C (The two quantities are equal):

The book's reasoning makes sense to me as it didn't dawn on me to use the "foil method" to factor the right hand side. From what the book says it looks like this makes sense. However when I PLUG IN the numbers it DOES NOT makes sense.

Here's another example in which Column B is in fact greater than Column A:
x = 4, y = 16
Column A = -2
Column B = 2
Column B is greater than Column A again!

IN CONCLUSION,
the discrepancy between my answer and the book's answer means one of three things are flawed. I'd like someone to help me figure out which.

1: my math is flawed and I came to the wrong conclusion each time I concluded that Column B is greater than Column A (from plugging in x=4, y=16; and plugging in x=2, y=3), when in fact plugging in these numbers should result in Column A = Column B.

2. The book used factoring incorrectly to conclude that Column A = Column B (but how?)

3. There is a flaw in the mathematical formula of foiling,
or [(x-y)(x-y)]^2 = x^2 - 2(xy) + y^2 ...but apparently not in every case, making this rule have exceptions to it, which I've never seen before in my life.

So where is the flaw? #1? #2? or #3?
• Posted by a hidden member.
This is a tough one, but maybe you are supposed to theoretically "compare" the equations without substituting particular values for x and y. Like, the value of x is x (whatever that may be) and the value of y is y (whatever that may be).

Does that make sense?
• Melos

Posts: 264
I think I have the answer, stay tuned for my long winded explanation and hopefully for those of you playing at home, you can follow along.

Short answer, you are correct the book is wrong. They forgot an absolute value sign.
• Posted by a hidden member.
Melos saidI think I have the answer, stay tuned for my long winded explanation and hopefully for those of you playing at home, you can follow along.

Short answer, you are correct the book is wrong. They forgot an absolute value sign.

Yeah, they probably should have put an absolute value sign around Column A. That way your -1 would have been 1.
• Posted by a hidden member.
Melos, oh thank you so much! This is good news for me. I'm looking forward to reading your thorough explanation! =)
• Melos

Posts: 264
What the book failed to include was an absolute value.

I can't remember all of my algebra proofs so I'll just do it by example like you did earlier.

You were correct in your first explanation, column B will always be a positive number the way the book wrote it.

Let's look at their explanation of the solution step by step. For ease, each step will be numbered as it is shown in the book.

1)

This one is easy to see, it's the same as column B and we know it is always positive when we substitute.

2)

Now let's look at the second part:

sqrt [ ( sqrt(x) - sqrt(y) ) ^ 2 ]

Let's call everything on the inside A:

A = ( sqrt(x) - sqrt(y) ) ^ 2

This little nugget will also always be positive, regardless of the values of x and y. The sqrt of a positive number must be a positive number, so step 2 must also be positive.

x = 4, y = 9

and begin with solving for A.

( sqrt(4) - sqrt(9) ) ^ 2 = (2 - 3) ^ 2 = ( -1) ^ 2 = 1

so A = 1

=> sqrt A = sqrt (1) = 1

Now let's switch the values of x and y like you did earlier:

x = 9, y = 4

( sqrt(9) - sqrt(4) ) ^ 2 = (3 - 2) ^ 2 = (-1) ^ 2 = 1

so A = 1

sqrt A = sqrt (1) = 1

Hopefully you can see, no matter what positive number combination you put into this expression, you get a positive number, just like with the original expression in Column B.

3)

This expression is the exact same as column A and where the book errors. When you take the square root of a square you will always always always receive a positive number. They do not simply cancel each other out.

so what it should look like is:

| sqrt(x) - sqrt(y) |

Let's use your example numbers yet again on the corrected expression:

x = 4, y = 9

| sqrt(4) - sqrt(9) | = | 2 - 3 | = | -1 | = 1

x = 9, y = 4

| sqrt(9) - sqrt(4) | = | 3 - 2 | = | 1 | = 1

So, in summary,

- you were correct

- the book did factor correctly, they just forgot the absolute value at the end

- foiling works, it can just be a pain to remember to do so sometimes.

I really hope this helps and doesn't make things worse. It's really hard to explain mathematical expressions via forum posts =/
• Posted by a hidden member.
Yup. Needs the absolute value (if Y > X).

Another way to look at it is to assume that they are equal.
In that case, you can multiply each side by itself.
Resulting in:
X -2 SqRt(XY) + Y =?= X -2 SqRt(XY) + Y

These are obviously equal. However, since we squared it, we lost clarity on the absolute value....
• hartfan

Posts: 1038
Since √(x²) = ± x

Therefore √((√x-√y)²) = ± (√x-√y)

So yes you have to factor in the absolute value. But otherwise the book is correct and so were you.
• Posted by a hidden member.
Sigh, this brings back bad memories of complex analysis...my i's always look like squiggles and my dot on the i looks more like an accent grave
• Vaughn

Posts: 1880
Math is Satan.
• Posted by a hidden member.
Being a grammar nazi at least has a practical application. This... just useless.
• Posted by a hidden member.
Lostboy saidBeing a grammar nazi at least has a practical application. This... just useless.

I assume you're joking...
• Posted by a hidden member.
hartfan saidSince √(x²) = ± x

Therefore √((√x-√y)²) = ± (√x-√y)

So yes you have to factor in the absolute value. But otherwise the book is correct and so were you.

How were you able to insert the square root sign, the exponent, and the plus/minus sign?
• Sk8Tex

Posts: 738
MenschPress said
hartfan saidSince √(x²) = ± x

Therefore √((√x-√y)²) = ± (√x-√y)

So yes you have to factor in the absolute value. But otherwise the book is correct and so were you.

How were you able to insert the square root sign, the exponent, and the plus/minus sign?

Start > Programs > Accesories > System Tools > Character Map

WOOT! I answered a math question! Go me
• Posted by a hidden member.
Just google "square root sign" and "plus/minus sign" and cut-paste.
God help us if Google goes down.
• Posted by a hidden member.
Lostboy saidBeing a grammar nazi at least has a practical application. This... just useless.

This, has to be a joke. It's a joke, right?
• Sk8Tex

Posts: 738
q1w2e3 saidJust google "square root sign" and "plus/minus sign" and cut-paste.
God help us if Google goes down.

Hold Alt and on the Numeric keypad press 0177, 0247 etc...Google was not always there
• Posted by a hidden member.