Question 683210: I will use R to represent a square root. Please help. thank you so much:)
1.(R5+R2)(-3R5+3R2)
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! There are multiple ways you can do this problem. I will show you the way that may be simpler:
[(sqrt)5 + (sqrt)2][-3(sqrt)5 + 3(sqrt)2]
[(sqrt)5 + (sqrt)2](-3)[(sqrt)5 - (sqrt)2]
(-3)[(sqrt)5 + (sqrt)2][(sqrt)5 - (sqrt)2].
Notice that I factored -3 from the second quantity. I can do that because they are not under the radicals. They are coefficients, and coefficients normally can be reduced, factored, or simplified in algebra I. Now, we must do distribution (looking at the square root quantities first):
[(sqrt)5 + (sqrt)2][(sqrt)5 - (sqrt)2]
(sqrt)5[(sqrt)5 - (sqrt)2] + (sqrt)2[(sqrt)5 - (sqrt)2]
(sqrt)25 - (sqrt)10 + (sqrt)10 - (sqrt)4
5 - 2 = 3.
Notice that this is basically FOIL. If you remember or know the special rules of binomial squares, conjugates (plus and minus), etc., then you can simply say that [(sqrt)5 + (sqrt)2][(sqrt)5 - (sqrt)2] is (sqrt)25 - (sqrt)4 or 3.
If you have factored the -3 out to begin with, you must not forget about it! Multiply 3 by -3 to get the final answer of -9.
Now I will show the longer way of doing it using "R" as the square root symbol besides (sqrt):
(R5 + R2)(-3R5 + 3R2)
R5(-3R5 + 3R2) + R2(-3R5 + 3R2)
-3R25 + 3R10 + -3R10 + 3R4
-3R25 + 3R4
-3(5) + 3(2)
-15 + 6 = -9.
The reason 3R10 and -3R10 can cancel out (or equal 0) is due to the fact that they both have the same radical. You can multiply different radicals, but you can only add or subtract radicals that are the same (in terms of simplifying or combining like terms).
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