need some help here:
Square root of 3x + 2 = 5 it says undo the radical)
and
Square root of x + 1 - Square root of x - 2 = 1, it says put radicals on opposite sides.
All help appreciated, and if you can please tell me how you did this, thank you.
2007-04-09 22:19:23 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Sq rt(3x+2)=5
Sq both sides: 3x + 2= 25
subtract 2
3x=23
divide by 3
x= 23/3
2007-04-10 12:47:26 · answer #1 · answered by Addie cain 3 · 0⤊ 0⤋
I suppose it is sqrt(3x+2)=5. Following that 3x+2 must be >=0, x>=-2/3.
Now let's find the solution:
sqrt(3x+2)=5
3x+2=25
x=23/3 which is greater than -2/3.
For the other part of the problem:
sqrt(x+1) -sqrt(x-2)=1
x+1>=0 means x>=-1 and x-2>=0 means x>=2, so x>=2
sqrt(x+1) = 1+sqrt(x-2)
x+1=1+2sqrt(x-2)+x-2
2=2sqrt(x-2)
sqrt(x-2)=1
x-2=1 means x=3
2007-04-09 22:41:46 · answer #2 · answered by no_nick 1 · 0⤊ 0⤋
The answer to the first question is 3. Here is my working
root 3x +2 = 5 (Now take 2 from each side)
root 3x = 3 (Now square both sides to get rid of the root)
3x = 9 (Now divide each side by 3)
x= 3
The second question are there any brackets anywhere???
2007-04-09 22:25:47 · answer #3 · answered by MattG83 1 · 0⤊ 0⤋
3x+2 = 25
sqrt(x+1) = 1+sqrt(x-2)
2007-04-09 23:00:47 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 0⤋