+839 Okay... So I have this problem, "Simplify the irrational number √32, then estimate it to two decimal places."
I got the second part which is 5.66, but I don't understand the first part. I have posted a question like this one before and tried to do what you guys said,but this is what I got.
32 is divisable by 2 so √16*√2.
But 16 is divisable by 2 so √8*√2
Butttttttt 8 is divisable by 2 also so √4*√2
Butttttttttttttttttttt 4 is divisable by 2 sooooooo √2*√2
and I don't think that is right.
Andddddddddd 2 is divisable by 2 soooooooooooooooo √1*√1!!!!!
I need help :P
+118608 32 is divisable by 2 so √16*√2.
It is not the 2 that is important here. It is the 16
32=16*2
If you want to simplify a surd you have to look for a squared number that goes into it.
squared numbers are (don't worry about 1, it is not helpful)
$$\\1^1=1, \;2^2=4,\;3^2=9,\;4^2=16\;etc\\\\
1,4,9,16,25,26,49,etc$$
$$\begin{array}{rlll}
\sqrt{32}&=&\sqrt{16\times2}\\\\
&=&\sqrt{16}\times \sqrt2&\\\\
&=&4\sqrt2&\\\\
\end{array}$$
so it is pulling out the square number that is the important bit!
+128460 Remember that we can write 32 as 16*2
So.......
√32 = √(16*2) = √16 * √2 = 4√2
+118608 32 is divisable by 2 so √16*√2.
It is not the 2 that is important here. It is the 16
32=16*2
If you want to simplify a surd you have to look for a squared number that goes into it.
squared numbers are (don't worry about 1, it is not helpful)
$$\\1^1=1, \;2^2=4,\;3^2=9,\;4^2=16\;etc\\\\
1,4,9,16,25,26,49,etc$$
$$\begin{array}{rlll}
\sqrt{32}&=&\sqrt{16\times2}\\\\
&=&\sqrt{16}\times \sqrt2&\\\\
&=&4\sqrt2&\\\\
\end{array}$$
so it is pulling out the square number that is the important bit!
