Question

Can someone show me the work to this?

Answer 1

To simplify the terms you execute a prime factorization (divide exactly by the smallest prime possible until you reach 1) which in the case of 320 would be: 320=2*2*2*2*2*2*5. Every factor which exists twice can be taken out of the root. Simplify the many 2s into 2 equal readable factors: 320=2*2*2*2*2*2*5=8*8*5

$\sqrt{320}=\sqrt{8*8*5}=8\sqrt{5}$

$\sqrt{162}=\sqrt{9*9*2}=9\sqrt{2}$

$\sqrt{108}=\sqrt{6*6*3}=6\sqrt{3}$

$\sqrt{125}=\sqrt{5*5*5}=5\sqrt{5}$

$\sqrt{63}=\sqrt{3*3*7}=3\sqrt{7}$

$\sqrt{432}=\sqrt{3*12*12}=12\sqrt{3}$

$\sqrt{112}=\sqrt{4*4*7}=4\sqrt{7}$

$\sqrt{98}=\sqrt{7*7*2}=7\sqrt{2}$

$\sqrt{12}=\sqrt{2*2*3}=2\sqrt{3}$

$\sqrt{726}=\sqrt{11*11*6}=11\sqrt{6}$

$\sqrt{300}=\sqrt{3*10*10}=10\sqrt{3}$

$\sqrt{8}=\sqrt{2*2*2}=2\sqrt{2}$

$\sqrt{27}=\sqrt{3*3*3}=3\sqrt{3}$

$\sqrt{18}=\sqrt{2*3*3}=3\sqrt{2}$

$\sqrt{196}=\sqrt{14*14}=14$

$\sqrt{30}=\sqrt{3*5*2}=\sqrt{3*10}$ and the end