There are a few problems from an assignment that I having some issues with. Could you please explain them?
1. √120x⁴y²/40x²y⁵ (just in case it isn't showing up clearly: 120x⁴y² "over" 40x²y⁵, and everything is under a square root sign)
2. 5x/ ⁴√80x (5x "over" the fourth root of 80x)
3. 1/ 1 +√2
Thanks so much for your help!
1. √120x⁴y²/40x²y⁵ (just in case it isn't showing up clearly: 120x⁴y² "over" 40x²y⁵, and everything is under a square root sign)
2. 5x/ ⁴√80x (5x "over" the fourth root of 80x)
3. 1/ 1 +√2
Thanks so much for your help!
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1. b/c they are both under a √ sign you can pretend like the √ sign doesn't exist and neither will I until the very end. So you're looking for ways to simplify, I'm going to go from numbers->x->y;
a- you have 120/40. 120 divided by 40= 3, so you can replace the 120 on top with a 3 and completely cross out the 40.
b- you have x^4 over x^2. When dividing variables of the same letter (i.e. both are x) you subtract the exponents and leave the variable the same. So here 4-2=2, giving you x^2 on top instead of x^4 and allowing you to completely cross out the x^2 on the bottom.
c- you have y^5 over y^5. Well they're the same thing so just cross them both out b/c by dividing them you get 1.
Now you bring back the √ sign.
Final Answer= √(3x^2)
2. So for this one you have plain 5x on top and ⁴√(80x) on the bottom. In this case you need to do something called rationalizing the denominator, in which you multiply both the top and bottom by something that will REMOVE the square root on the bottom and bring it to the top instead. So this is what I would do;
a- multiply top and bottom by ⁴√(80x)^3, thereby canceling out the square root on the bottom and leaving you with 80x, and instead bringing ⁴√(80x)^3 to the top.
So so far you have; (5x * ⁴√(80x)^3)/ 80x
b- Now I''ll cancel out the 5x and divide the 80x on the bottom by 5x, thus giving me;
⁴√((80x)^3)/16.
3. 1/ (1+√2). In this case you need to multiply the top and bottom by the conjugate of the denominator. By conjugate I mean b/c we are given (1+√2), I multiply it by (1-√2) and turn the bottom into 1-2 = -1, in order to cancel out the √ on the bottom and bring it to the top, giving me;
(1-√2)/ -1 ---> √(2) - 1
a- you have 120/40. 120 divided by 40= 3, so you can replace the 120 on top with a 3 and completely cross out the 40.
b- you have x^4 over x^2. When dividing variables of the same letter (i.e. both are x) you subtract the exponents and leave the variable the same. So here 4-2=2, giving you x^2 on top instead of x^4 and allowing you to completely cross out the x^2 on the bottom.
c- you have y^5 over y^5. Well they're the same thing so just cross them both out b/c by dividing them you get 1.
Now you bring back the √ sign.
Final Answer= √(3x^2)
2. So for this one you have plain 5x on top and ⁴√(80x) on the bottom. In this case you need to do something called rationalizing the denominator, in which you multiply both the top and bottom by something that will REMOVE the square root on the bottom and bring it to the top instead. So this is what I would do;
a- multiply top and bottom by ⁴√(80x)^3, thereby canceling out the square root on the bottom and leaving you with 80x, and instead bringing ⁴√(80x)^3 to the top.
So so far you have; (5x * ⁴√(80x)^3)/ 80x
b- Now I''ll cancel out the 5x and divide the 80x on the bottom by 5x, thus giving me;
⁴√((80x)^3)/16.
3. 1/ (1+√2). In this case you need to multiply the top and bottom by the conjugate of the denominator. By conjugate I mean b/c we are given (1+√2), I multiply it by (1-√2) and turn the bottom into 1-2 = -1, in order to cancel out the √ on the bottom and bring it to the top, giving me;
(1-√2)/ -1 ---> √(2) - 1