Question

Who ever sucks at Math like me come here!!!

Answer 1

I seen a problem I am going to do it, tell me If I am correct.... OK?

Answer 2

\[\sqrt{2r}=2+\sqrt{r-2}\]
\[\sqrt{2r}-\sqrt{r-2}=2\]
\[2r-2\sqrt{2r(r-2)}+r-2=4\]

not done yet....

Answer 3

I am having comp probs hold on...

Answer 4

\[3r-2\sqrt{2r^2-4}+ r-2=4\]
\[2\sqrt{2r^2-4}=3r-6\]

am doing good so far?

Answer 5

\[8r^2-16=9r^2-32r+52\]
\[-r^2+32r-68=0\]
\[r^2-32r+68=0\]
is this good so far?
\[(r-16)^2=188\]
\[r= 16 ± 2\sqrt{47}\]

right?

Answer 6

Look what I found! @LifeIsADangerousGame

Answer 7

this is one of your yesterday's problems isn't it?

Answer 8

@SolomonZelman, mistake after squaring both sides look over your work.....

Answer 9

haha yes, this is my problem from yesterday

Answer 10

You need to square both sides Solomon

Answer 11

what did you get as answer there do you remember?

Answer 12

what I am getting is \[r^2+28r+4=0\]

Answer 13

can't factor

Answer 14

\[r^2+28=-4\]
\[(r+14)^2=-4+196\]
\[r=-14± 32\sqrt{3}\]

Answer 15

r approximately equal to 41.43 or -69.43

Answer 16

\[3a - 4 \le 5\]

Answer 17

@Victoria_Cakes20, do the following: (Step by Step)

1) add 4 to both sides
2) divide both sides by 3

do this carefully and correctly (what's your answer?)

Answer 18

(2r)^1/2 = 2 + (r - 2)^1/2
(2r)^1/2 - 2 = (r - 2)^1/2

Square both sides
2r - 4(2r)^1/2 + 4 = r - 2

Take the radical to the right and everything else to the left

2r + 4 - r + 2 = 4(2r)^1/2
r + 6 = 4(2r)^1/2

Square both sides

r^2 + 12r + 36 = 16(2r) = 32r
r^2 + 12r + 36 - 32r = 0
r^2 - 20r + 36 = 0
r^2 - 2r - 18r + 36 = 0
r(r - 2) - 18(r - 2) = 0
(r - 18)(r - 2) = 0

r = 18 or r = 2

Answer 19

Yes, how didn't I figured this out?

Answer 20

And it works perfectly for each.