Asked by Micah
This looks like a trick question
(sqrt3+5 sqrt6)(sqrt3-5 sqrt6)
Do I sq 3 or do I add 3 and 5 first?
I think I have to do the same to both sides.
(sqrt3+5 sqrt6)(sqrt3-5 sqrt6)
Do I sq 3 or do I add 3 and 5 first?
I think I have to do the same to both sides.
Answers
Answered by
drwls
Here's a clue: (a+b) (a-b) = a^2 - b^2
Your problem is in that form. a and b can be anything.
Just take the difference of the squares of the two terms, sqrt 3 and 5 sqrt6. Then you will have the answer. It will be a negative integer.
Your problem is in that form. a and b can be anything.
Just take the difference of the squares of the two terms, sqrt 3 and 5 sqrt6. Then you will have the answer. It will be a negative integer.
Answered by
Micah
3sqrt is 9
6sq rt is 36
I add 5 to one and subtract 5 from the other? 50 and 55, then I am supposed to multiply? 50x55=2750? Or is the difference -5?
6sq rt is 36
I add 5 to one and subtract 5 from the other? 50 and 55, then I am supposed to multiply? 50x55=2750? Or is the difference -5?
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