Question

simplify the square root of 20 puls the square foot of 40

Answer 1

can anyone help not sure how to solve?

Answer 2

helo, welcome to openstudy!

Answer 3

\[\sqrt {20}+\sqrt{40}\]

simplify the numbers by factoring

\[20 = 4\times5=2^2\times5 \\40=\]

Answer 4

can you factor 40 into prime factors ?

Answer 5

yes it would be 4*10=2^3*5 right?

Answer 6

Right.

Answer 7

i am not really sure what there is to "simplify" here
you can write \(\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\times \sqrt{5}=2\sqrt{5}\) which puts it in simplest radical form

Answer 8

so we can write\[\sqrt {20}+\sqrt{40}=\sqrt{2^2\times5}+\sqrt{2^3\times5}\]

now if there are terms under a square root that are squared, we can take them out, and drop the square ,

\[=2\sqrt{5}+\sqrt{2^3\times5}\]

for the second radical we need to break up the 2^3 so we can extract the square term , 2^3 = (some square term) times (something else)

Answer 9

can you simplify the second term @iun45 ?

Answer 10

2^2*2 right

Answer 11

good,

Answer 12

now take the square term out,

Answer 13

so we get 2

Answer 14

do you mean =2√5+2√(2×5) ?

Answer 15

2^2√5+2^2√5*2 ?

Answer 16

the ^2 bit cancels with the √ bit when you bring the term out from under the racial


so

\[\quad\sqrt{2^2\times 5}+\sqrt{2^2\times 5\times2}\\=\sqrt{2^2}\times\sqrt{ 5}+\sqrt{2^2}\times \sqrt{5\times2}\\=2\times\sqrt{ 5}+2\times \sqrt{5\times2}\\=2\sqrt{ 5}+2 \sqrt{5\times2}\]

Answer 17

i have used this for simplifying the square roots\[\sqrt{ab}=\sqrt a\sqrt b\]

use it again to factor the last term,

Answer 18

finally factor out the common factors of each term

\[ab+abc=ab(1+c)\]

Answer 19

ok now i'm lost