Question

how do you solve this equation (20-r)^1/2=r

Answer 1

naynay, you need to square both sides so you can get access to the r under the square root.

So\[20-r=r^2 \rightarrow r^2+r-20=0\]which is quadratic. This can now be factored as\[(r+5)(r-4)=0\]so \[r=-5\]or\[r=4\]

Answer 2

how do you solve this (6b)^1/2=(8-2b)^1/2

Answer 3

One sec.

Answer 4

ok

Answer 5

You have to remove the square roots. You do this by squaring. Whatever you do to one side, you must do to the other, so,\[\sqrt{6b}=\sqrt{8-2b} \rightarrow 6b=8-2b \rightarrow 8b=8 \rightarrow b=1\]

Answer 6

ok so b will equal 1

Answer 7

yes

Answer 8

oh ok i see what you did

Answer 9

ok this one i really dont get -3=(37-3n)^1/2-n

Answer 10

Have you become a fan yet?

Answer 11

no i will

Answer 12

You're always looking to isolate your variable. Here I would add n to both sides and then square both sides, so\[n-3=\sqrt{37-3n} \rightarrow (n-3)^2=37-3n\]Then expand the left hand side\[n^2-6n+9=37-3n \rightarrow n^2-3n-28=0\]which you can then solve by factoring.

Answer 13

where did you get 37 from

Answer 14

You gave it to me in your question:

-3=(37-3n)^1/2-n

Answer 15

oh ok im sorry didnt look twice =)

Answer 16

np

Answer 17

thank you a lot with helping me because im really not all that good in math

Answer 18

You're welcome. All you need is practice.

Answer 19

Things will start to click.

Answer 20

yeah i have a friend of mine comin to help me twice a week so hopefully it work =)

Answer 21

I'm sure it will. Good luck, and you can use this site too for help.

Answer 22

yeah i like this site a lot it does help

Answer 23

can you help me with this (-3-4x)^1/2-(-2-2x)^1/2=1

Answer 24

\[\sqrt{-3-4x}-\sqrt{-2-2x}=1\]Okay, for ones like this, it's best to move one of the square roots to the other side and then square both sides:\[\sqrt{-3-4x}=1+\sqrt{-2-2x}\]square both sides:\[-3-4x=\left( 1+\sqrt{-2-2x} \right)^2\]Now the hard part is to expand the right-hand side\[\left( 1+\sqrt{-2-2x} \right)^2=1+2\sqrt{-2-2x}+\left( \sqrt{-2-2x} \right)^2\]\[=1+2\sqrt{-2-2x}+(-2-2x)\]\[=-1-2x+\sqrt{-2-2x}\]

Answer 25

ok so this girl helped me and we got it right =)

Answer 26

Okay, so now looking at this, we have the problem of the square root again. This is because of the expansion we had to do earlier BUT we can get rid of it using the same technique as before - move everything that has nothing to do with the square root to the other side, and then square both sides.\[-3-4x=-1-2x-\sqrt{-2-2x}\]becomes\[-2-2x=-\sqrt{-2-2x}\]which is the same as\[\sqrt{-2-2x}=2+2x \rightarrow -2-2x=(2+2x)^2\]\[-2-2x=4+8x+4x^2 \rightarrow 4x^2+10x+6=0\]i.e.\[2x^2+5x+3=0 \rightarrow (x+2)(x+3)=0 \rightarrow x=-2,-3\]

Answer 27

Well, I just thought I'd finish it.