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If x=3/4 and y=2/5, what is the value of [#permalink]
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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
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Originally posted by ajit257 on 26 Feb 2011, 16:36.
Last edited by walker on 27 Oct 2012, 04:25, edited 2 times in total.
Edited the question.



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If x=3/4 and y=2/5, what is the value of [#permalink]
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26 Feb 2011, 17:22
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.
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Re: tricky fractions [#permalink]
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26 Feb 2011, 17:28
Awesome! Bunuel...thanks a ton ! apologies about rewording the question.
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Re: tricky fractions [#permalink]
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26 Feb 2011, 18:54
sqrt(y1)^2 > You cannot calculate square root of negative number.
B is correct as Bunuel said.



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Re: tricky fractions [#permalink]
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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



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Re: tricky fractions [#permalink]
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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 \(2/51=3/5=3/5.\)
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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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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



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



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Re: tricky fractions [#permalink]
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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.
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Re: tricky fractions [#permalink]
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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 \(\sqrt{x^2 +6x +9}\sqrt{y^2 2y +1}=\sqrt{(x+3)^2}\sqrt{(y1)^2}\)
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Re: tricky fractions [#permalink]
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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.
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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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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?
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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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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
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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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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



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



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Re: If x=3/4 and y=2/5, what is the value of [#permalink]
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01 Aug 2016, 22:07
This question is asking us to recognize that the expression under the roots are perfect squares. we can write x^2+ 6*x+ 9 as (x+3)^2 and the other one as (y1)^2. sqroot((x+3)^2) = sqroot ((15/4)^2) = 15/4 sqroot((y1)^2) = sqroot((0.41)^2) = sqroot((0.6)^2). Now since we are first squaring and then taking root the result is 0.6. As Bunuel pointed out the sqroot(x^2) is equivalent to x. We don't have to necessarily work out the absolute value function as long as we do squaring and then taking the root. The answer then is 15/46/10 which is choice B.



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