How do you evaluate #(\sqrt{(3-1)^{2}+(8-0)^{2}#?

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Answer 1 · 2017-10-19T00:21:29

#sqrt(68)=2*sqrt(17)#

#sqrt((3-1)^2+(8-0)^2)#

The way to simplify these expressions is to work inside out. First, I'll do the subtraction in the parentheses.

#sqrt((2)^2+(8)^2)#

Now, I'll compute each square.

#sqrt(4+64)#

Next, I'll add.

#sqrt(68)#

Then, I'll try to find perfect squares that are factors of the number in the radical. In this case, we can see that #68=4*17=2^2*17#. Thus

#sqrt(68)=sqrt(2^2*17)=sqrt(2^2)*sqrt(17)=2*sqrt(17)#