SOLUTION: Can you please help me solve this radical equaqtion? {{{sqrt(x-1)+x=3}}} This problem did not come from a text book, it came from a worksheet. I began by isolating the radical:

Algebra ->  Radicals -> SOLUTION: Can you please help me solve this radical equaqtion? {{{sqrt(x-1)+x=3}}} This problem did not come from a text book, it came from a worksheet. I began by isolating the radical:      Log On


   



Question 179343: Can you please help me solve this radical equaqtion? sqrt%28x-1%29%2Bx=3
This problem did not come from a text book, it came from a worksheet.
I began by isolating the radical: sqrt%28x-1%29=-x%2B3
Then I squared each side to eliminate the radical: sqrt%28x-1%29%5E2=%28-x%2B3%29%5E2
Which gave me: +x-1=-x%5E2%2B9+
Then I moved everything to one side to give me a quadradtic equation:
x%5E2%2Bx-10=0
This is where I got stuck, I don't see how this can be factored, so I tried using the quadratic equation and just got confused from there. Can you help me please??!!???

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Can you please help me solve this radical equaqtion? sqrt%28x-1%29%2Bx=3
This problem did not come from a text book, it came from a worksheet.
I began by isolating the radical: sqrt%28x-1%29=-x%2B3
Then I squared each side to eliminate the radical: sqrt%28x-1%29%5E2=%28-x%2B3%29%5E2
Which gave me: +x-1=-x%5E2%2B9+

There's your error!

%28-x%2B3%29%5E2 IS NOT -x%5E2%2B9!

When there are two terms added to be squared, you must
write it in parentheses TWICE and multiply it by itself, 
using FOIL. Also you must remember that when you square 
a negative, such as %28-x%29 you get a positive %22%22%2Bx%5E2,
not -x%5E2:

Therefore when you got to

sqrt%28x-1%29%5E2=%28-x%2B3%29%5E2

the next step should have been

x-1=%28-x%2B3%29%28-x%2B3%29

Then you must use "FOIL":

x-1=%28-x%29%28-x%29+%2B+%28-x%29%283%29+%2B+%283%29%28-x%29+%2B+%283%29%283%29

matrix%281%2C3%2Cx-1%2C+%22=%22%2C%22%22%2Bx%5E2+-+3x+-+3x+%2B+9%29

You don't need to write that %22%22%2B%22%22. I just
put it there to remind you that when you square
a negative you get a positive.

matrix%281%2C3%2Cx-1%2C%22=%22%2Cx%5E2-3x-3x%2B9%29

x-1=x%5E2-6x%2B9

Then move everything right and you get:

0=x%5E2-7x%2B10

or if you like the 0 on the right, just swap sides:

x%5E2-7x%2B10=0

Then it factors nicely as

%28x-5%29%28x-2%29=0

Then you get the solutions x=5 and x=2

But you must check these, because when you square
both sides or multiply both sides by a variable,
you may get extraneous solutions:

Checking x=5
sqrt%28x-1%29%2Bx=3
sqrt%285-1%29%2B5=3
sqrt%284%29%2B5=3
2%2B5=3
7=3

That doesn't check, so x=5 is not a 
genuine solution, but is called "extraneous".

Checking x=2
sqrt%28x-1%29%2Bx=3
sqrt%282-1%29%2B2=3
sqrt%281%29%2B2=3
1%2B2=3
3=3

That checks. So x=2 is the only solution

Edwin