Part A:
To solve the equation sqrt(2x + 10) - 1 = x, we need to isolate the radical term and then square both sides of the equation to eliminate the radical.
1. Add 1 to both sides of the equation:
sqrt(2x + 10) = x + 1
2. Square both sides of the equation:
(sqrt(2x + 10))^2 = (x + 1)^2
This simplifies to:
2x + 10 = x^2 + 2x + 1
3. Subtract 2x + 10 from both sides of the equation:
0 = x^2 + 1 - (2x + 10)
0 = x^2 - 2x - 9
4. Rearrange the equation in standard quadratic form:
x^2 - 2x - 9 = 0
5. Solve the quadratic equation. We can either factor it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-(-2) ± sqrt((-2)^2 - 4(1)(-9))) / (2(1))
Simplifying this expression gives:
x = (2 ± sqrt(4 + 36)) / 2
x = (2 ± sqrt(40)) / 2
x = (2 ± 2sqrt(10)) / 2
6. Simplify the expression for x:
x = 1 ± sqrt(10)
Therefore, the solutions to the equation sqrt(2x + 10) - 1 = x are x = 1 + sqrt(10) and x = 1 - sqrt(10).
Part B:
To determine which solution is extraneous, we need to substitute both x-values back into the original equation and check if they satisfy it.
Let's substitute x = 1 + sqrt(10) into the original equation:
sqrt(2(1 + sqrt(10)) + 10) - 1 = 1 + sqrt(10)
Simplifying the left-hand side of the equation gives:
sqrt(2 + 2sqrt(10) + 10) - 1 = 1 + sqrt(10)
This further simplifies to:
sqrt(12 + 2sqrt(10)) - 1 = 1 + sqrt(10)
In this case, it is difficult to determine if the equation is satisfied. However, let's substitute x = 1 - sqrt(10) into the original equation:
sqrt(2(1 - sqrt(10)) + 10) - 1 = 1 - sqrt(10)
Simplifying the left-hand side of the equation gives:
sqrt(2 - 2sqrt(10) + 10) - 1 = 1 - sqrt(10)
This further simplifies to:
sqrt(12 - 2sqrt(10)) - 1 = 1 - sqrt(10)
We can see that both solutions satisfy the original equation, and therefore, neither of them is extraneous.
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%0D%0A%0D%0Asqrt(2x + 10) - 1 = x%0D%0A%0D%0A %0D%0A%0D%0APart A) Solve the equation and show all necessary steps. (3 points)%0D%0A%0D%0A2x+10−−−−−−√−1=x%0D%0A2%0D%0A%0D%0A+%0D%0A10%0D%0A−%0D%0A1%0D%0A=%0D%0A%0D%0A %0D%0A%0D%0APart 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