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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
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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
_________________

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??
_________________

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 \(\sqrt{x^2}=|x|\).
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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) ...

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.

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, \(|-\frac{3}{5}| = -(-\frac{3}{5}) = \frac{3}{5}\)

When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root . Thats why i didn't use mod. What am i missing here?

When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root . Thats why i didn't use mod. What am i missing here?

Precisely because the square root sign cannot give negative result you should use modulus sign here. x + 3 depending on x can be negative, while |x + 3| cannot.

I think this is a high-quality question and I agree with explanation. I think you can make this question a bit harder if you add 83/20 as an answer choice for people like me who forgot to take the absolute value into the consideration.