Official Solution:

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. \(\frac{87}{20}\)
B. \(\frac{63}{20}\)
C. \(\frac{47}{20}\)
D. \(\frac{15}{4}\)
E. \(\frac{14}{5}\)

\(\sqrt{(x+3)^2} - \sqrt{(y-1)^2} = |x+3| - |y-1| = | \frac{3}{4} + 3| - |\frac{2}{5}-1| = \frac{15}{4}-\frac{3}{5}=\frac{63}{20}\)


Answer: B
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Because \(\sqrt{x^2}=|x|\). Check for more Absolute value tips: absolute-value-tips-and-hints-175002.html


Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hope this helps.
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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
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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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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.
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|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.
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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.

Check for more the discussion above, specifically the following post: m06-183699.html#p1465193

Hope it helps.
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

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

Thank you. Edited the typo.
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