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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
A. 87/20
B. 63/20
C. 47/20
D. 15/4
E. 14/5
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26 Feb 2011, 17:22
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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
Note that \(\sqrt{x^2}=|x|\)
\(\sqrt{x^2 +6x +9}-\sqrt{y^2 -2y +1}=\)
\(= \sqrt{(x+3)^2}-\sqrt{(y-1)^2}=\)
\(= |x+3|-|y-1|=\)
\(= |\frac{3}{4}+3|-|\frac{2}{5}-1|=\)
\(= |\frac{15}{4}|-|-\frac{3}{5}|=\)
\(= \frac{15}{4}-\frac{3}{5}=\)
\(= \frac{63}{20}\)
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.
A. 87/20
B. 63/20
C. 47/20
D. 15/4
E. 14/5
Note that \(\sqrt{x^2}=|x|\)
\(\sqrt{x^2 +6x +9}-\sqrt{y^2 -2y +1}=\)
\(= \sqrt{(x+3)^2}-\sqrt{(y-1)^2}=\)
\(= |x+3|-|y-1|=\)
\(= |\frac{3}{4}+3|-|\frac{2}{5}-1|=\)
\(= |\frac{15}{4}|-|-\frac{3}{5}|=\)
\(= \frac{15}{4}-\frac{3}{5}=\)
\(= \frac{63}{20}\)
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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