Question

4squ2-squ32+squ16 simply @phi

Answer 1

\[4 \sqrt2-\sqrt{32}+\sqrt{16}=4 \sqrt2-\sqrt{4^2 \times 2}+\sqrt{4^2}\]
\[=4 \sqrt2-4\sqrt{ 2}+4=4\]

Answer 2

@jdoe0001

Answer 3

yeap

Answer 4

looks like you are getting the hang of it.

Answer 5

I got 4squ2+8squ2-4squ2 but its not a choice

Answer 6

4squ2-squ32+squ16

I got 4squ2+8squ2-4squ2 but its not a choice

4 sqr(2) is ok
- sqr(32) is - 4 sqr(2). which matches your -4squ2

how did you get 8 squ2 ?

Answer 7

16*2

Answer 8

which term did you start with ?

Answer 9

the first one

Answer 10

I am lost. If we start at the beginning
4squ2-squ32+squ16 simplify

if we write this nicely
\[ 4 \sqrt{2} - \sqrt{32} + \sqrt{16} \]

the first term \( 4 \sqrt{2}\) is already simplified (you can not factor the 2 inside the square root)

now the second term
\[ - \sqrt{32} \]
first, can you factor the 32 ?

Answer 11

32= 2*16 (which if we know 16 is a perfect square is good enough) but to continue
= 2* 2* 8= 2*2*2 *4 = 2*2*2*2*2
we have 2 multiplied by itself 5 times.
How many pairs are there ? group into pairs (like noah's ark):
(2*2) * (2*2) * 2
so
\[ \sqrt{32} = \sqrt{ (2 \cdot 2)(2 \cdot 2)2}\]