SOLUTION: Solve. The square root of x-1 = x-3 I got as far as x=x^2-6x+10 but don't know what to do next. Please explain. Thank you.

Algebra ->  Radicals -> SOLUTION: Solve. The square root of x-1 = x-3 I got as far as x=x^2-6x+10 but don't know what to do next. Please explain. Thank you.      Log On


   

Question 87473: Solve.
The square root of x-1 = x-3
I got as far as x=x^2-6x+10 but don't know what to do next. Please explain.
Thank you.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x-1%29+=+x-3
x-1+=+%28x-3%29%5E2
x-1=+x%5E2+-6x%2B9

Now, since this is a quadratic equation, you must set it equal to zero by subtracting x and adding +1 to each side:
0=x%5E2+-7x+%2B10

This factors! By the way in radical equations, they ALWAYS factor!!

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

You MUST check the answers, since you squared both sides of the equation!!

sqrt%28x-1%29+=+x-3
Check: x=5
sqrt%285-1%29+=+5-3
sqrt%284%29=+2 It checks!!

Check: x=2
sqrt%282-1%29+=+2-3
sqrt%281%29=+-1 It does NOT check!!
You must reject the value of x=2.

Final answer is x=5 only!

R^2 Retired from SCC