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Question 693879: How do your simplify: A.√15v * √3b
B. √5(2√10 + √2x) C. (2-√6)(1 + √6)
Please show steps.
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! 1. √15v * √3b
When multiplying radicals, we want to multiply everything underneath the radicals together and keep them under the radical. We multiply whole numbers together and we multiply the radicals together. So,
√15v * √3b
√(15v*3b) (multiply everything under radicals)
√45bv (keeping everything under the radical).
Now we need to simplify the number underneath the radical. 45 can be written as a product of 9 * 5, which 9 can be square rooted to get 3 outside of the radical. In other words,
√45bv
√[(9)(5)(bv)] (9*5*bv = 45bv, so we can write it this way)
3√5bv (we can take the square root of any factor of a number that is already underneath a radical as long as it is not a decimal).
The simplified answer is 3√5bv. This answer is equivalent to √45bv and gives the same answer no matter what b and v equal (as long as they remain the same in both expressions). The factor 9 of 45 can be square rooted and brought outside of the radical by multiplication/radical rules. The 5 can't be square rooted into a whole number, so it remains under the radical. B and v remain underneath the radical because they do not have a square sign on them. Variables must have a square sign on them to be brought out from the radical. To check whether or not 3√5bv is the equivalent form of √45bv, we can square 3√5bv and see if equals 45bv (not worrying about the radical). That is,
(3√5bv)^2
(3√5bv)(3√5bv) (definition of squaring terms)
(3*3)(√5bv*√5bv) (multiply whole numbers together and radicals together)
9(√5bv)^2 (definition of squaring terms)
9(5bv) (square signs cancel out radicals)
45bv (multiplication).
Since the square of 3√5bv equals 45bv, and the square of √45bv equals 45bv, 3√5bv and √45bv must be equivalent. The first is just the simplified version of the latter.
2. √5(2√10 + √2x)
Here we have distributive property to worry about. The rules still stay the same. So,
√5(2√10 + √2x)
√5(2√10) + √5(√2x) (distributive property)
2(√5*√10) + (√5*√2x) (multiplying radicals)
2(√50) + (√10x) (multiplying everything under radicals)
2√50 + √10x (remove parentheses won't change expression in this case)
2√(25*2) + √10x (50 can be written as a product of 25 and 2)
2(5)√2 + √10x (the square root of 25 is 5; bring it out and multiply it to any whole number already outside of the radical)
10√2 + √10x (multiply 2 and 5).
It is dependent on your teacher which way you write it, but the general idea is that the simplified expression is 10√2 + √10x. If you were to calculate this expression into a decimal and if you did the same for the original expression √5(2√10 + √2x), the decimal would be the same. To check ourselves to see if 10√2 is the simplified radical of 2√50, we do the same thing as we did in the first problem (we must make sure that the square of this term and the square of the original term 2√50 are the same):
(10√2)^2
(10√2)(10√2)
(10*10)(√2*√2)
100(√2)^2
100(2)
200.
Is this equal to the square of 2√50?
(2√50)^2
(2√50)(2√50)
4(√50)^2
4(50)
200.
It is, so we have simplified 2√50 correctly.
3. (2 - √6)(1 + √6)
For this one, we just use FOIL as if this was simple binomial multiplication. So,
(2 - √6)(1 + √6)
2(1 + √6) + (-√6)(1 + √6) (FOIL)
2 + 2√6 + (-√6) + (-√36) (when we multiply a whole number and radical together, we just write them side by side like 2√6).
2 + 2√6 - 1√6 - 1√36 (we can't forget that there is a -1 in front of √6 and √36 when there is a negative sign outside of the radical)
2 + 1√6 - √36 (subtract 2√6 and 1√6 because they share the same radical)
2 + 1√6 - 6 (the square root of 36 is 6)
-4 + 1√6 (combine like terms 2 and -6)
-4 + √6 (you don't have to write the 1 in front of √6 in final answer).
To check ourselves, we can simply calculate the original expression and the final expression to a decimal and see if they match. If they do, then we've done it correctly (assuming we've simplified as far as we can).
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