ajit257 wrote:
If x=3/4 and y=2/5 , what is the value of sqrt(x+3)^2 - sqrt(y-1)^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 \(\sqrt{x^2}=|x|\)
\(\sqrt{x^2 +6x +9}-\sqrt{y^2 -2y +1}=\sqrt{(x+3)^2}-\sqrt{(y-1)^2}=|x+3|-|y-1|=|\frac{3}{4}+3|-|\frac{2}{5}-1|=|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}=\frac{63}{20}\)
Answer: B.
As for your question:
The point here is that
square root function 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.
How did you take \(|-\frac{3}{5}| as \frac{3}{5}\) because \(\sqrt{(y-1)^2}\)= |y-1| and now |y-1| can have two values (y-1), if y-1>0 => y>1 or -(y-1) if y-1<0 => y<1 .... and here y is given as 2/5 (y<1), that means value of \(\sqrt{(y-1)^2}\) = |y-1| = -(y-1)
\(|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-(-\frac{3}{5}) = \frac{15}{4}+\frac{3}{5} = \frac{87}{20}\)
please, let me know if I am doing anything wrong here.