Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The point here is that square root function cannot give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Hi Bunuel,

In this step: \(|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}\)

How did you take \(|-\frac{3}{5}| as \frac{3}{5}\) because \(\sqrt{(y-1)^2}\)= |y-1| and now |y-1| can have two values (y-1), if y-1>0 => y>1 or -(y-1) if y-1<0 => y<1 .... and here y is given as 2/5 (y<1), that means value of \(\sqrt{(y-1)^2}\) = |y-1| = -(y-1) that means \(|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-(-\frac{3}{5}) = \frac{15}{4}+\frac{3}{5} = \frac{87}{20}\)

please, let me know if I am doing anything wrong here.

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Hi Bunuel,

In this step: \(|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}\)

How did you take \(|-\frac{3}{5}| as \frac{3}{5}\) because \(\sqrt{(y-1)^2}\)= |y-1| and now |y-1| can have two values (y-1), if y-1>0 => y>1 or -(y-1) if y-1<0 => y<1 .... and here y is given as 2/5 (y<1), that means value of \(\sqrt{(y-1)^2}\) = |y-1| = -(y-1) that means \(|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-(-\frac{3}{5}) = \frac{15}{4}+\frac{3}{5} = \frac{87}{20}\)

please, let me know if I am doing anything wrong here.

thanks

-K

Without any connection to where it came from, \(|-\frac{3}{5}|=\frac{3}{5}\). Absolute value expresses distance and can never be negative. \(|y-1|\) can never have two values. \(|8|=8\), while \(|-8|=-(-8)=8\). So, \(|y-1|=|2/5-1|=1-2/5=3/5\) or \(|2/5-1|=|-3/5|=3/5.\) _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

If \(x<0\), then \(|x|=-x\). \(|-8|=8=-(-8)\). If \(x>0\), then \(|x|= x. \,\,|0|=0.\) \(|x|\) means the distance on the number line between \(x\) and \(0.\) Distance between \(5\) and \(0\) is the same as the distance between \(-5\) and \(0.\) So, when the number is positive, absolute value of it is the number itself. When the number is negative, the absolute value of the number is that number without the minus sign. There is no mathematical operation of "drop the sign of the negative number." But if we multiply a negative number by \(-1\), we get that number without its negative sign. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Hi Bunuel,

I have got lil confused here.. if x^2=25 => x=sqrt(25) => x=5 right ? then why x^2=25 => +5,-5 if the above can be written as x^2=25 => x=sqrt(25) => x=5 ? Please help me understand this !!

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Hi Bunuel,

I have got lil confused here.. if x^2=25 => x=sqrt(25) => x=5 right ? then why x^2=25 => +5,-5 if the above can be written as x^2=25 => x=sqrt(25) => x=5 ? Please help me understand this !!

Thanks a ton in advance

\(x^2=25\) has two solutions: \(x=\sqrt{25}=5\) and \(x=-\sqrt{25}=-5\). But, \(\sqrt{25}=5\), because square root function cannot give negative result.

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Bunuel, another dubious question. . If I substitue the value of x and y here, then I dont have to take the mod thing because root of square of a number is always positive and the value results in 87/20 \(\sqrt{x^2 +6x +9}-\sqrt{y^2 -2y +1}=\sqrt{(x+3)^2}-\sqrt{(y-1)^2}\) _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

The point here is that square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

About \(\sqrt{x^2}=|x|\): from above we have that \(\sqrt{x^2}\geq{0}\). 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 \(\sqrt{x^2}=|x|\).

Hope it's clear.

Bunuel, another dubious question. . If I substitue the value of x and y here, then I dont have to take the mod thing because root of square of a number is always positive and the value results in 87/20 \(\sqrt{x^2 +6x +9}-\sqrt{y^2 -2y +1}=\sqrt{(x+3)^2}-\sqrt{(y-1)^2}\)

There is nothing wrong with this questions as well! The answer is 63/20. Check your math. _________________

Square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

Square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

Re: If x=3/4 and y=2/5, what is the value of [#permalink]

Show Tags

10 Jul 2014, 09:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x=3/4 and y=2/5, what is the value of [#permalink]

Show Tags

12 Sep 2015, 10:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Post your Blog on GMATClub We would like to invite all applicants who are applying to BSchools this year and are documenting their application experiences on their blogs to...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...