X+2 = (square root) 3x+10 is of the form x+a = (square root) bx+c, where a, b, and c are all positive integers and. 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, where a, b, and c are all positive integers and 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

Buster

26 June, 01:20

X+2 = (square root) 3x+10 is of the form x+a = (square root) bx+c, where a, b, and c are all positive integers and. 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, where a, b, and c are all positive integers and 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.

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Answers (2)

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Judith Sweeney · Answer 1

Hello there.

X+2 = (square root) 3x+10 is of the form x+a = (square root) bx+c, where a, b, and c are all positive integers and. 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, where a, b, and c are all positive integers and 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.

Answer for both: no solutions

Dante Webster · Answer 2

X + 2 = √ (3x + 10)

In the form:

x + a = √ (bx + c)

If we let x = 7

7 + a = √ (7b + c)

Squaring both sides

7 + 14a + a² = 7b + c

Assume a = 1

7 + 14 + 1 = 7b + c

22 = 7b + c

Assume a = 2

39 = 7b + c

There are no solutions to the the two equations.