M06-02

Why do we use absolute signs in this problem??

Hi Bunuel,
Agreed that as per the theory & property its \(\sqrt{x^2}=|x|\)
But why can't we cancel out the square root with the power square?
Because i did it as \(\sqrt{(x+3)^2} - \sqrt{(y-1)^2}\)
after cancelling the roots ----> (x+3)-(y-1)---> \(x-y+4 = 3/4 - 2/5 + 4\)
and answer came out to be 87/20.
Please help

Because \(\sqrt{(x+3)^2}=|x+3|\), not x+3 and \(\sqrt{(y-1)^2} = |y-1|\), not y-1. Check for more here: absolute-value-tips-and-hints-175002.html

I checked the absolute tips and \(\sqrt{x^2}=|x|\) is given as one of the properties
But how the property is derived or holds is not mentioned.
Is this property holds because in GMAT only positive values are considered for even roots and thus \(\sqrt{x^2}=|x|\)
right??

About \(\sqrt{x^2}=|x|\):

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hi Bunuel,
I understand your explanation.
I wanted to apply the mod and solve this problem, However, we also know that..."When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
So, the solution to this problem can go either way 87/20 (considering the positive roots only) or 63/20 ( with the mod approach) ...

PEMDAS
parentheses comes first. we cannot jump operations and square/take the square root of everything before solving for what is in parenthesis.
y-1 results in a negative expression. After squaring it, it gives us a positive expression. Moreover, after squaring everything, the square root will be as well a positive number. I did not do the way bunuel explained, yet I got to the right answer.

sqrt[(x+3)^2] = sqrt[(3/4+12/4)^2] = sqrt[(15/4)^2]. this will result in 15/4.
now let's take y
sqrt[(y-1)^2] = sqrt[(2/5-5/5)^2] = sqrt[(-3/5)^2]. now, if we square -3/5 we get 9/25 which is a positive number. sqrt of 9/25 is 3/5.

we got 15/4 - 3/5
multiply first by 5/5 and second by 4/4, the result is:
(15*5)/20 - (3*4)/20. extend this and get: (75-12)/20.
the result is 63/20.

Hello Bunuel,
As per Official Guide, |x| is defined to be x if x >= 0 and -x if x < 0. Here it is |x+3| - |y-1| = |15/4| - |-3/5|.
Now (y-1) = -3/5 which is less than 0.

So shouldn't it be |15/4| - |-3/5| = (15/4) - (-3/5) = (15/4) + (3/5) = 87/20 ?

Kindly correct me if I'm making a mistake. Thanks.

|x| = -x if x <= 0, yes. But the point behind this is that absolute value cannot be negative, for example |-5| = 5. If x is negative, then |x| = -x = -negative = positive. Here, \(|-\frac{3}{5}| = -(-\frac{3}{5}) = \frac{3}{5}\)

Hope it's clear.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

This question has its own classs

GMATClub

I think this is a high-quality question and I agree with explanation. I was impressed by the approach you took, excellent and quite useful question.

An alternative approach: Hello, everyone. Sometimes I like to challenge myself by working through these tough GMAT Club problems mentally—without writing down anything—so that I develop a better habit of consulting the given information rather than working from memory and perhaps falling into a trap answer. In this case, I did recognize that the quadratics under the square roots were perfect squares, but because algebra is more difficult to keep track of mentally than numbers, I worked with the numbers only and solved the question in under two minutes. I will outline the process below.


Step #1: Convert each quadratic to a binomial—

\(x^2+6x+9=(x+3)^2\) and

\(y^2-2y+1=(y-1)^2\)

Step #2: Substitute the given values of x and y into each binomial and convert, when necessary, into an improper fraction—

\(\frac{3}{4}+3=\frac{15}{4}\) and

\(\frac{2}{5}-1=-\frac{3}{5}\)

Step #3: Understand that the positive root of these values, when squared, will simply be the positive value itself. For instance,

\(\sqrt{(\frac{15}{4})^2}=\frac{15}{4}\) and

\(\sqrt{(-\frac{3}{5})^2}=\frac{3}{5}\)

Thus, there is no reason to actually square each fraction and then take the square root.

Step #4: Solve as you would a regular difference involving fractions, finding a common denominator—

\(\frac{15}{4}-\frac{3}{5}\)

\(\frac{75}{20}-\frac{12}{20}\)

\(\frac{63}{20}\)

The answer must be (B).


I hope that helps. Yes, there are still a few steps to navigate, but with a few fundamentals in place, the problem can be solved efficiently, with or without jotting down numbers. (Note: I would highly encourage writing down intermediary numbers during practice or on the actual exam.)

Good luck with your studies.

- Andrew

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

absolute value tend to be ignored and one can make silly mistake easily. brilliant question. must revisit this even if you get it right.