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Answer:
- x -3 = √(2x +2)
- x = 1 is extraneous
- bx +c > 0
Step-by-step explanation:
(a) The form of the problem guarantees there will be an extraneous solution. We only need to find values of a, b, c that make the equation true for x=7.
When x=7, we have ...
7 -a = √(7b +c)
Since a, b, c are positive integers, and b > 1, the smallest possible value we can have under the radical is 7·2+1 = 15. The largest possible value we can have under the radical is (7-1)^2 = 36, which gives several choices for b and c. Since all of these numbers are integers, the number under the radical must be a perfect square. That is, it must be one of {16, 25, 36}.
We choose 7b+c = 16, so we must have b=2, c=2. Then 7-a = √16 = 4, which means a=3. Our equation is ...
x -3 = √(2x +2)
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The extraneous solution arises when x -3 = -√(2x +2). (Note the minus sign in front of the radical.) If we square both sides of this equation, we get ...
x^2 -6x +9 = 2x +2
x^2 -8x +7 = 0 ⇒ (x -7)(x -1) = 0 ⇒ x = 1 or 7
The extraneous solution is x=1.
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(b) For the equation to have real solutions, the value of bx+c cannot be negative. The square root function does not give real values for negative arguments. That is, we require ...
bx +c > 0
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Additional comment
For a, b, c all positive integers and b > 1, the curve defined by the radical will begin on the negative x-axis and pass through the second quadrant on its way to end behavior in the first quadrant.
The line x-a will have a negative y-intercept and a positive slope. It will always intersect the radical curve in the first quadrant. In short, no additional restrictions need to be placed on a, b, c in order to guarantee a real number solution.
