~*MATH HELP pleas!!!???*~?

Question

adding and subtracting radical expressions

4√6+√7-6√2+4√7

and

3√7-2√28

please show all work.
thanks so much!!!:]

Answer 1

PLEASE REMEMBER that you can only add or subtract radical terms where the radical is the same in the terms. Sometimes, a little housekeeping helps in this respect.

In the first one, we can add terms with sqrt(7), but not the other two(although we can do some factoring). So the answer would be
5 sqrt(7) + 2 (2 sqrt(6) -3 sqrt(2) )

In the second problem , we appear to have the same bad-news situation, but not really, since we can "break-up" sqrt(28) into 2 sqrt(7). So we wind up with - sqrt(7)

Answer 2

There is a very important radical and square roots rule to bear in mind:

When there is a factor beside the square root, you can square it and multiply it by the number beneath the square root.

For example in 4√6:

Square 4 to get 16. Multiply 16 by 6 to get 96 and there you will have 4√6 = √96

Now do this to the rest of them with factors:

√96 + √7 - √72 + √112

Now you can add them and subtract them like normal numbers:

√96 + √7 - √72 + √112 = √143




The second is the same too. Just square 3 and multiply it by 7 beneath the square root and do the same with 2√28:

3√7 - 2√28 = √63 - √112 = -√49

-√49 has a perfect root which is 7 so the result will be -7.


Good luck.