2. The equation x+2 = sqrt (3x+10) is of the form x+a=sqrt (bx+c), where a, b, and c are all positive integers and b > 1. Using this equation as a model, create your own equation that has extraneous solutions. (a) Using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt (bx+c) where a, b, and c are all positive integers and b > 1 such that 7 is a solution. (Hint: Substitute 7 for x, and choose a value for a. Then square both sides so you can choose a, b, and c that will make the equation true.) (b) Solve the equation you created in Part 2a. (c) If your solution in Part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. If your solution in Part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.

Question

Mathematics

Jamison Boyd

18 January, 02:20

2. The equation x+2 = sqrt (3x+10) is of the form x+a=sqrt (bx+c), where a, b, and c are all positive integers and b > 1. Using this equation as a model, create your own equation that has extraneous solutions. (a) Using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt (bx+c) where a, b, and c are all positive integers and b > 1 such that 7 is a solution. (Hint: Substitute 7 for x, and choose a value for a. Then square both sides so you can choose a, b, and c that will make the equation true.) (b) Solve the equation you created in Part 2a. (c) If your solution in Part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. If your solution in Part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.

+5

Answers (1)

Know the Answer?

Martin Baker · Answer 1

(a). We substitute 7 for x to find that 7+a=sqrt (7b+c). Letting a arbitrarily equal 1, we see that sqrt (7b+c) = 8, so 7b+c=64. One possible solution to this equation is b=9 and c=1, for the equation x+1=sqrt (9x+1).

(b). Squaring the equation, we see that x^2+2x+1=9x+1. Subtracting 2x+1, we see that x^2=7x. If x=0, 0^2=7*0, so x=0 is one solution, while otherwise, we can divide by x to see that x=7. Thus, our solutions are 7 and 0.

(c). Doing this task is tantamount to finding a different solution to 7b+c=64. One such solution is b=8 and c=8, for the equation x+1=sqrt (8x+8).