Question

( √2 - √3 )² =

Answer 1

please help

Answer 2

\[(\sqrt2-\sqrt3)^2=(\sqrt2-\sqrt3)(\sqrt2-\sqrt3)\]Use the distributive property to simplify the right side.

Answer 3

*Let me know if you need another step

Answer 4

yes, I dont know how distribute because they are in parentheses. What exactly do I do to multiply these?

Answer 5

You multiply each term in the first set of parentheses by each term in the second set of parentheses. Let's look at another example. Let's use (5-3)a. When we simplify this using the distributive property, we get this:
\[(5-3)a=5a-3a\]
Does this^ bit make sense? I can break it down more if it will help.

Answer 6

yes

Answer 7

so do i just add the square roots together when I distribute?

Answer 8

Your question might mean a couple different things, but you could think of the example like this:
\[(5+(-3))a=5a+(-3a)\]
This would mean your problem would look like this:
\[(\sqrt2+(-\sqrt3))(\sqrt2+(-\sqrt3))\]

Answer 9

So you're multiplying the individual square roots together, but then you add the terms at the end. It'll look like this:
\[=\sqrt2(\sqrt2-\sqrt3)+-\sqrt3(\sqrt2-\sqrt3)\]

Answer 10

It's actually a bit easier to see with letters:
\[(a+-b)(\sqrt2-\sqrt3)=a(\sqrt2-\sqrt3)+-b(\sqrt2-\sqrt3)\]

Answer 11

Im just confused on how I would multiply a \[\sqrt{2}\] by a \[\sqrt{3}\].

Answer 12

Ah, okay. Are you okay to this point:
\[(\sqrt2-\sqrt3)(\sqrt2-\sqrt3)=\sqrt2\sqrt2-\sqrt2\sqrt3-\sqrt2\sqrt2+\sqrt3\sqrt3\]

Answer 13

how did u get two 2sqr2's?

Answer 14

Oh, whoops, that was a mistype. That should read
\[\sqrt2\sqrt2-\sqrt2\sqrt3-\sqrt2\sqrt3+\sqrt3\sqrt3\]

Answer 15

Sorry about that :(

Answer 16

its ok,im good up to that point. Now what do I do with that equation

Answer 17

Alright, let's look at this:
\[\sqrt2\sqrt3=\sqrt{(2)(3)}=\sqrt6\]

Answer 18

As long as it's just multiplication and division and they all have radicals (with the same root), it doesn't matter whether you work the inside or the outside

Answer 19

oh ok

Answer 20

Good explanation :)

Answer 21

@Jhannybean thanks :)

Answer 22

alright so how do I end up with whole numbers in the final answer? Where do they come from?

Answer 23

Alright, so we have four parts. Here's the first:
\[\sqrt2\sqrt2=\sqrt4=2\]
What are the other three?

Answer 24

sqr2sqr3=sqr6

Answer 25

Yep, but make sure it's negative

Answer 26

remember: \((\color{red}{a}-\color{blue}{b})^2 = \color{red}a^2 -2\color{red}a\color{blue}{b}+\color{blue}{b}^2 \implies (\color{red}{\sqrt{2}}-\color{blue}{\sqrt{3}})^2\)

Answer 27

@Jhannybean how did you make them different colors?

Answer 28

right click on my latex > show math as > TeX commands

Answer 29

woah woah woah, how did u get -2ab? Maybe thats the step Im missing in this.

Answer 30

because that is the format when you are taking the difference of two squares :)

Answer 31

Keep working it and you'll see a second -sqrt(6) pop out

Answer 32

Awesome, I learned something today :)

Answer 33

Mmhmm, once you have completed @jabberwock's method, go back to my post and see if you can follow the format! :D

Answer 34

Sorry for interrupting~

Answer 35

Thanks @Jhannybean

Answer 36

sqr-6 sqr-6 sqr9 sqr4

Answer 37

Ah, careful. -sqrt6, not sqrt(-6).

-sqrt(6) is to the left of zero. sqrt(-6) isn't a real number.

Now we add all those terms together:
\[(-\sqrt6) + (-\sqrt6) + \sqrt9 +\sqrt4\]
Can you simplify any of those square roots?

Answer 38

simplify we get: -sqr6 -sqr6 sqr3 and sqr2

Answer 39

oh wait -sqr6 -sqr6 3 and 2

Answer 40

Good :)

Answer 41

So if you add all those together, you get \[5+(-\sqrt6)+(-\sqrt6)\]What do you get if you add \[-\sqrt6+(-\sqrt6)\]

Answer 42

2sqr6?

Answer 43

Good, but negative

Answer 44

so it will be 5-sqr6?

Answer 45

So finally,
\[(\sqrt2-\sqrt3)^2=5-2\sqrt6\]

Answer 46

1 hour later... lol thanks for your help man. Much obliged :)

Answer 47

Lol, sorry. But I don't think you'll have much problem with others that look like it. Best of luck.

Answer 48

thanks

Answer 49

No worries :)