Radicals help please?!?

Question

1. √112x^4-√7x^4

2. √18-√50

3. (3√x-√y)(√x-4√y)

4. 12-15√6+8√6-10√36

Answer 1

1. √(112x^4) - √(7x^4) =
4x²√7 - x²√7 =
3x²√7

2. √18 - √50 = 3√2 - 5√2 = -2√2

3. (3√x - √y)(√x - 4√y) = 3x - 13√(xy) + 4y

4. 12 - 15√6 + 8√6 - 10√36 =
12 - 7√6 - 60 =
-48 - 7√6

Answer 2

The first one, I'm not sure whats in sqrts. Is it sqrt(11)(2x^4) or sqrt(112)(x^4) or a number of other things?

2) Find two numbers that when multiplyed together, equal the number in the radical. So with 18 theres 1 and 18. Well neither of those can be simplified. 3 and 6. Can't take sqrt of either of those. 9 and 2. Hey! sqrt(9)= 3. Can't take sqrt(2) so we'll leave it as so. Now we have 3sqrt(2). Since we have 2 under the sqrt in that, we want sqrt(2) in the other one. 25 and 2= 50. sqrt(25)=5 and leave sqrt(2) as that.
So we have 3sqrt(2)-5sqrt(2)
Pretend sqrt(2) is a varible when adding or subtracting.
3sqrt(2)- 5sqrt(2)= -2sqrt(2)

3) FOIL!
3sqrt(x^2)+ 12sqrt(xy)- sqrt(xy)- 4sqrt(y^2)
take sqrt of (x^2) and (y^2)
3x+ 12sqrt(xy)- sqrt(xy)- 4y
3x+ 11sqrt(xy)- 4y

4) Add -15 and 8.
12- 7sqrt(6)-10(6)
12- 7sqrt(6)- 60
32- 7sqrt(6)

Answer 3

1) sqrt. (112) = sqrt. (7 x 16) = (sqrt. 7) x 4
Substitute and take sqrt.7 x^4 common.
This gives
sqrt. 7 (4 - 1) = 3 sqrt. 7

2) 3 sqrt.2 - 5 sqrt.2 = -2 sqrt.2

3) 3√x (√x-4√y) -√y (√x-4√y) = 3x - 12√xy - √xy + 4y = 3x - 13√xy + 4y

4) 12-15√6+8√6-10(6) = 12 - 7√6 - 60 = -48 - 7√6
If you want, you may take -1 common.

Hope that helped.

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