SOLUTION: Solve for , where is a real number. (under a radical) -x+6 + (under a different radical) x-5 = 1 (If there is more than one solution, separate them with commas.)

Question 545537: Solve for , where is a real number.
(under a radical) -x+6 + (under a different radical) x-5 = 1
(If there is more than one solution, separate them with commas.)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(under a radical) -x+6 + (under a different radical) x-5 = 1
----
sqrt(-x+6) + sqrt(x-5) = 1
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Square both sides to get:
(-x+6) + (x-5) + 2sqrt(-x^2+11x-30) = 1
---------
2sqrt(-x^2+11x-30) = 0
---
sqrt(-x^2+11x-30) = 0
---
-x^2+11x-30 = 0
---
x^2-11x+30 = 0
---
Factor:
(x-6)(x+5) = 0
-----
Acceptable answer:
x = 6
============
Cheers,
Stan H.
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