Question

I need help with one more problem if someone will help me. I'll put it below:

Answer 1

\[(2\sqrt{5}+3\sqrt{7})^{2}\]

Answer 2

(a+b)^2 = (a+b)(a+b)

Answer 3

So: \[(2\sqrt{5} + 3\sqrt{7})(2\sqrt{5} + 3\sqrt{7})\]

Answer 4

?

Answer 5

Yes

Answer 6

foil

Answer 7

Thanks @optiquest !

Answer 8

@optiquest So I got this:

Answer 9

Ugh it messed up. It was supposed to say\[2\sqrt{5}(2\sqrt{5}+3\sqrt{7})+3\sqrt{7}(2\sqrt{5} + 3\sqrt{7})\]

Answer 10

I dont think you foiled right

Answer 11

Yeah I do it differently than my teacher does. I usually get the right answer, but I'm not sure now. :/

Answer 12

\[(2\sqrt{5})^2 + 3\sqrt{7}\times2\sqrt{5} + (3\sqrt{7})^2\]

Answer 13

\[2^{2}\sqrt{5}^{2}+3\times7^{1/2}\times2\times5^{1/2}+3^{2}\sqrt{7}^{2}\]

Answer 14

\[4\times5+3\times 7^{.5}\times2\times5^{.5}+9\times7\]

Answer 15

see above

Answer 16

Where did all of those numbers come from?

Answer 17

\[20+63+6\times(7*5)^{.5}\]

Answer 18

which ones

Answer 19

The 5 and 3 and 9

Answer 20

im not sure where you are looking

Answer 21

Okay I understand every part except for the exponents. What is the .5 from?

Answer 22

you can just rewrite a root at a number raised to 1/2 or .5 they are equivalent

Answer 23

@Hero do think you could help me out?

Answer 24

\[x ^{.5} = x ^{1/2} = \sqrt{x}\]

Answer 25

\[2\sqrt{5}(2\sqrt{5}+3\sqrt{7})+3\sqrt{7}(2\sqrt{5} + 3\sqrt{7})\]
youre distribution is fine, notice we have 2 more distributions, finish it out
\[2\sqrt{5}(2\sqrt{5})+2\sqrt{5}(3\sqrt{7})+3\sqrt{7}(2\sqrt{5}) + 3\sqrt{7}(3\sqrt{7})\]
\[4\sqrt{25}+6\sqrt{35}+6\sqrt{35} + 9\sqrt{49}\]