Question

Consider equations of the form x - a= sqrt(bx+c) , where a, b, and c are all positive integers and b > 1.

(a) Create an equation of this form that has 7 as a solution and an extraneous solution. Give the extraneous solution.

(b) What must be true about the value of bx+c to ensure that there is a real number solution to the equation? Explain.


Only answer if you can explain this to me correctly. Do not answer for points please.

Answer 1

Explanation:

x - a =√(bx+c)

a) for 7 as a solution, the equation could be :

x - 1 = √(4x +8) , where a=1, b=4 and c = 8.

and the solution as follows :

(x-1)² = 4x+8

x²-2x+1=4x+8

x²-2x-4x+1-8=0

x²-6x-7 = 0

(x-7) (x+1) =0

x= 7 or x= -1

so, the extraneous solution is -1

b) in order to get the real numbers solutions, the value of (bx+c) should be ≥ 0.

Answer 2

9514 1404 393

Answer:

• x -3 = √(2x +2)
• x = 1 is extraneous
• bx +c > 0

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

_____

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.