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Hero Group TitleBest ResponseYou've already chosen the best response.0
\[\text{Let} \space \sqrt{2} = x \space \text{then} \space 6\sqrt{2}  3\sqrt{2} = 6x  3x = 3x = 3\sqrt{2}\]
 2 years ago

imagreencat Group TitleBest ResponseYou've already chosen the best response.0
Basically, you are just going to combine like terms. And when one term has a sqrt(2), and there's a value before it, we can take 3sqrt(2) + 2sqrt(2) to be equal to 5sqrt(2). It's like saying 3 cows + 2 cows = 5 cows. the sqrt(2) is like the label of each term. Actually, like what Hero said above. Now, you may be confused with the second and third numbers because they contain two labels now. Don't mix them, because sqrt(y) is not equal to the sqrt(x) of the sqrt(z). These are three different labels. Just leave them that way. Like the expressing 3x + 2y = 5z is left that way.
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Do you still need help?
 2 years ago

BazBendell Group TitleBest ResponseYou've already chosen the best response.0
I have no idea what either of you said...
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
When you add or subtract the terms with root, the best way is factorize the root out!
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Do you understand my explanation?
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
For example: 6√2  3√2 = ( 6 3 ) *√2 = 3 √2
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Of course, only the terms have the same root can be factored!
 2 years ago

BazBendell Group TitleBest ResponseYou've already chosen the best response.0
Not a clu what that means.
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
6√2  3√2 Which part do you see the same?
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I hope @BazBendell isn't a first grader trying to understand 7th grade work.
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
@BazBendell would you prove that Hero is completely wrong ;)
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
The only way he could prove me wrong is by understanding the material somehow ;)
 2 years ago
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