Part A:
sqrt(2x + 10) - 1 = x
Add 1 to both sides:
sqrt(2x + 10) = x + 1
Square both sides to get rid of the square root:
2x + 10 = (x + 1)^2
2x + 10 = x^2 + 2x + 1
Subtract 2x and 10 from both sides:
0 = x^2 - 9
x^2 - 9 = 0
(x - 3)(x + 3) = 0
x = 3 or x = -3
Therefore, the solutions are x = 3 and x = -3.
Part B:
The extraneous solution is x = -3. We can see this by back-substituting x = -3 back into the original equation:
sqrt(2(-3) + 10) - 1 = -3
sqrt(4) - 1 = -3
2 - 1 = -3
1 = -3
Since 1 does not equal -3, x = -3 is an extraneous solution and is not valid for the given equation.
Note: This question has 2 parts. You must fully answer both parts in order to receive full credit. When typing your work, use "sqrt" to represent the radical symbol, followed by parentheses to enclose anything under that radical. For example, you would type the equation in Part A as
sqrt(2x + 10) - 1 = x
show all your work
Part A) Solve the equation and show all necessary steps. (3 points)
2x+10−−−−−−√−1=x
2
𝑥
+
10
−
1
=
𝑥
Part B) One of your solutions is extraneous, which means it is not a valid solution to the problem. Which solution is extraneous? Explain how you know. (2 points)
1 answer