## allie_bear22 Group Title et one year ago one year ago

1. shrinifores

can you give it in an equation form ? this is so confusing.

2. phi

you took the square root of perfect squares. if you can find a pair inside a square root, you can "pull one out" example: sqrt(3*3) = 3 (take out the pair of 3's, and replace with a single 3 outside the square root) for letters: sqrt(x^3 * x^3) you get x^3

I know what your asking but I do not know hot to do the problem... sorry sweety :)

4. phi

$\sqrt {32 x^7}$ break up the number 32 into 2*16= 2*2*8= 2*2*2*4= 2*2*2*2*2 now look for pairs: (2*2) * (2*2) * 2 we have two pairs of 2's and one left over. pull out each pair, and replace it with a single 2 $\sqrt {32 x^7} = \sqrt{ (2 \cdot 2) \cdot (2 \cdot 2) \cdot 2 \cdot x^7}= 2 \sqrt{ (2 \cdot 2) \cdot 2 \cdot x^7} = 2 \cdot 2\sqrt{ 2 \cdot x^7} = 4 \sqrt{ 2 \cdot x^7}$

5. phi

for the x^7 it is easiest if we know that x^7 means x times itself 7 times x*x*x * x*x*x * x and group 3 x's together to get a pair of (x*x*x) $4 \sqrt{2 x^7} = 4\sqrt{2 \cdot ( x\cdot x \cdot x) \cdot (x\cdot x \cdot x) \cdot x}$ pull out the pair, and replace it with just once copy outside the square root $4 \cdot (x\cdot x \cdot x) \sqrt{2 x}$ normally you would use exponents to make it look prettier: $4 x^3\sqrt{2x}$