Question 1120210: Find all values of x for which sqrt(x-3) > (x-5)
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
I wrote my first solution to this problem . . .
Then I saw that is was not exactly correct. Therefore, I deleted it.
Then I developed my second solution, which is in this post. Hope, it is correct.
Fist of all, since the condition does not state an opposite, I will assume that only positive branch
of the square root  is under consideration.
Then the solution consists of thee steps.
1. The domain of the left side square root is the set of real numbers x >= 3.
2. The set of those "x" what make the right side negative is
x - 5 < 0 ====> x < 5.
Thus the set 3 <= x < 5 is the part of the solution set to the given inequality, since the left side then exist and is positive,
while the right side is negative.
3. Now we will assume that x >= 5. Then both sides of the given inequality are positive.
Square both sides. Since both sides were positive, we have a valid inequality in the domain x >= 5
x- 3 >  ,
x-3 > x^2 - 10x + 25,
0 > x^2 - 11x + 28,
x^2 - 11x + 28 < 0,
Factor left side
(x-4)*(x-7) < 0
The solution is this interval 4 < x < 7.
Thus the final solution is the union of these sets 3 <= x < 5 and 4 < x <7.
This union is the interval 3 <= x < 7.
Thus the final ANSWER is THIS: 3 <= x < 7.
Let us check if we get the full set solution correctly.
I made this check graphically. Below please find the plots of the functions y = (red line) and the function y = x-5 (green line).
The solution set is those values of "x" on the x-axis, where the first plot is above the second one.
Plot of original functions y = (red) and y = x-5 (green)
Visual check confirms that the interval 3 <= x < 7 is the full set of solutions.
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