Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 28 Aug 2010
Posts: 257

If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
26 Feb 2011, 16:36
7
This post received KUDOS
29
This post was BOOKMARKED
Question Stats:
45% (02:31) correct
55% (01:20) wrong based on 1327 sessions
HideShow timer Statistics
If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ? A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit
Last edited by walker on 27 Oct 2012, 04:25, edited 2 times in total.
Edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 39744

If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
26 Feb 2011, 17:22
13
This post received KUDOS
Expert's post
18
This post was BOOKMARKED
ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 \(\sqrt{x^2}=x\). Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: tricky fractions [#permalink]
Show Tags
29 Sep 2012, 01:54
154238 wrote: Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 => +5,5if 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. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
01 Aug 2016, 22:33
ajit257 wrote: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?
A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 You can approximate too if perfect square doesn't come to mind. \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) \(\sqrt{x^2 +6x +9}\) > 3/4 square is 9/16 which is slightly more than 0.5 3/4 of 6 is 4.5 So .5 + 4.5 + 9 is 14 and its square root is slightly less than 4, say about 3.7 \(\sqrt{y^2 2y +1}\) > 2/5 square is 4/25 which is very small. 2/5 of 2 is 0.8 So 0.8 + 1 = 1/5 and its square root is about 1/2.2 which is 0.5 So \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1} = 3.7  0.5 = 3.2\) (slightly more than 3) Only option (B) is slightly more than 3.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 28 Aug 2010
Posts: 257

Re: tricky fractions [#permalink]
Show Tags
26 Feb 2011, 17:28
Awesome! Bunuel...thanks a ton ! apologies about rewording the question.
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit



Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 900

Re: tricky fractions [#permalink]
Show Tags
26 Feb 2011, 18:54
sqrt(y1)^2 > You cannot calculate square root of negative number.
B is correct as Bunuel said.



Manager
Joined: 27 Dec 2011
Posts: 70

Re: tricky fractions [#permalink]
Show Tags
26 Sep 2012, 07:01
Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 \(\sqrt{(y1)^2}\)= y1 and now y1 can have two values (y1), if y1>0 => y>1 or (y1) if y1<0 => y<1 .... and here y is given as 2/5 (y<1), that means value of \(\sqrt{(y1)^2}\) = y1 = (y1) 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



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: tricky fractions [#permalink]
Show Tags
26 Sep 2012, 10:29
kartik222 wrote: Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 \(\sqrt{(y1)^2}\)= y1 and now y1 can have two values (y1), if y1>0 => y>1 or (y1) if y1<0 => y<1 .... and here y is given as 2/5 (y<1), that means value of \(\sqrt{(y1)^2}\) = y1 = (y1) 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. \(y1\) can never have two values. \(8=8\), while \(8=(8)=8\). So, \(y1=2/51=12/5=3/5\) or \(2/51=3/5=3/5.\)
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 27 Dec 2011
Posts: 70

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
26 Sep 2012, 10:36
hi Evajager, if that the case then can please tell me why bunuel is considering "x" here: http://gmatclub.com/forum/ifx0thenrootxxis100303.html#p773743thanks, K



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
26 Sep 2012, 10:58
kartik222 wrote: hi Evajager, if that the case then can please tell me why bunuel is considering "x" here: http://gmatclub.com/forum/ifx0thenrootxxis100303.html#p773743thanks, K 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.



Intern
Joined: 09 Sep 2012
Posts: 33
Location: United States

Re: tricky fractions [#permalink]
Show Tags
29 Sep 2012, 01:48
Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 => +5,5if the above can be written as x^2=25 => x=sqrt(25) => x=5 ? Please help me understand this !! Thanks a ton in advance



Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 520
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: tricky fractions [#permalink]
Show Tags
12 Feb 2013, 06:50
Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}\)
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595
My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: tricky fractions [#permalink]
Show Tags
12 Feb 2013, 08:09
Sachin9 wrote: Bunuel wrote: ajit257 wrote: If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2  sqrt(y1)^2 ?
a. 87/20 b. 63/20 c. 47/20 d. 15/4 e. 14/5 Bunuel: Please can you clarify this concept. Thanks x = sqrt(25) > x = 5 x^2 = 25 > x = 5 Please don't reword the questions. Original question is: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 Note that \(\sqrt{x^2}=x\) \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}=x+3y1=\frac{3}{4}+3\frac{2}{5}1=\frac{15}{4}\frac{3}{5}=\frac{15}{4}\frac{3}{5}=\frac{63}{20}\) Answer: B. As for your question: 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 \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}\) There is nothing wrong with this questions as well! The answer is 63/20. Check your math.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 520
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
14 Feb 2013, 01:03
thanks bunuel. you r right but i don't understand the below thing in math: why is root [(3)square] not equal to 3 Why is it equal to 3 and not 3?
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595
My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
14 Feb 2013, 02:26



Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 520
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
14 Feb 2013, 02:44
Bunuel wrote: Sachin9 wrote: thanks bunuel. you r right but i don't understand the below thing in math:
why is root [(3)square] not equal to 3
Why is it equal to 3 and not 3? That's explained in my post above: ifx34andy25whatisthevalueof110071.html#p880130Square 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. Thus, \(\sqrt{(3)^2}=\sqrt{9}=3\). Or, applying \(\sqrt{x^2}=x\): \(\sqrt{(3)^2}=3=3\). Hope it's clear. amazing, thanks bro.. btw i like your sword in your pic
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595
My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Manager
Joined: 04 Mar 2013
Posts: 88
Location: India
Concentration: General Management, Marketing
GPA: 3.49
WE: Web Development (Computer Software)

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
03 Jul 2013, 07:36
ajit257 wrote: If \(x=\frac{3}{4}\) and \(y=\frac{2}{5}\), what is the value of \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}\) ?
A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5 y1 turns to 1y as y is less than 1 , once this is recognized its done x+ 3  (1y) = x plus y plus 2 = 23/20 plus 2



Senior Manager
Joined: 13 May 2013
Posts: 469

Re: If x=3/4 and y=2/5, what is the value of [#permalink]
Show Tags
03 Jul 2013, 15:08
If x=3/4 and y=2/5 What is the value of (sq. rt)(x^2+6x+9)  (sq. rt)(y^22y+1)
A. 87/20 B. 63/20 C. 47/20 D. 15/4 E. 14/5
(sq. rt)(x^2+6x+9) (sq. rt)(x+3)*(x+3) (sq. rt)(x+3)^2 x+3 (sq. rt)(y^22y+1) (sq. rt)(y1)*(y1) (sq. rt)(y1)^2 y1 SO x+3y1
3/4 + 3  2/51 15/4  3/5 75/20  12/20 75/20  12/20 =63/20 (B)



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16022

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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16022

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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If x=3/4 and y=2/5, what is the value of
[#permalink]
12 Sep 2015, 10:29



Go to page
1 2
Next
[ 21 posts ]




