Hi, I just have two Alg 2 questions that I need help on. If you can help, thanks! (Also, I have answers, I just need to know am I accurate and are they right or?)
1. Which statement best select the solution(s) of the equation?
1/x-1 + 2/x = x/x-1
Options:
There are two solutions: x = 1 and x = 2.
There is only one solution: x = 2.
The solution x = 1 is an extraneous solution.
There is only one solution: x = 1.
The solution x = 2 is an extraneous solution.
There is only one solution: x = 2.
The solution x = 0 is an extraneous solution.
My answer-- The second choice. Am I right?
2. Which statement best select the solution(s) of the equation?
Sqrt 2x-4-x+6=0
Options:
There is only one solution: x = 10.
The solution x = 0 is an extraneous solution.
There are two solutions: x = 4 and x =10.
There is only one solution: x = 10.
The solution x = 4 is an extraneous solution.
There is only one solution: x = 4.
The solution x = 10 is an extraneous solution.
My answer--The third choice. Am I right?
+128408 1/x-1 + 2/x = x/x-1 subtract the first fraction from both sides
2/x = x/ [ x - 1 ] - 1/ [ x - 1 ]
2/x = [ x - 1 ] / [ x - 1 ] cross-multiply
2 [x - 1] = x [ x - 1 ]
2x - 2 = x^2 - x rearrange as
x^2 - 3x + 2 = 0 factor as
(x - 2) ( x - 1) = 0 set each factor to 0 and solve for x and we find that
x = 2 or x = 1
The second solution makes an original denominator 0 we must reject that solution
So...the only solution is that x = 2 ....you are correct !!!
+2440 The square root sign is ambiguous in your second question. Can you tell me which equation you mean?
A) \(\sqrt{2x}-4-x+6=0\)
B) \(\sqrt{2x-4}-x+6=0\)
C) \(\sqrt{2x-4-x}+6=0\)
D) \(\sqrt{2x-4-x+6}=0\)
Without knowing this, it will be very difficult to help you out.
+2440 Ok, thanks for replying. One way to check if your solution is right is to plug it in and see if the resulting equation is indeed a true one. Let's do that!
\(\sqrt{2x-4}-x+6=0\)
Let's plug in x=4 and x=10 and see if we get a true statement.
Check x=4
| \(\sqrt{2*4-4}-4+6=0\) | Let's evaluate what is inside the radical to start and determine whether or not this is a true statement. |
| \(\sqrt{8-4}-4+6=0\) | |
| \(\sqrt{4}-4+6=0\) | The radicand, 4, is a perfect square, so this can be simplified further. |
| \(2-4+6=0\) | This is elementary addition and subtraction now. |
| \(4=0\) | This is an false. statement, so this solution does not satisfy the original. |
Check x=10
| \(\sqrt{2*10-4}-10+6=0\) | Ok, simplify inside the radical just like before determine if true just like before. |
| \(\sqrt{20-4}-10+6=0\) | |
| \(\sqrt{16}-10+6=0\) | Just like in the previous problem, the number underneath the square root is a perfect square, so the presence of them go away. |
| \(4-10+6=0\) | |
| \(0=0\) | This is a true statement, so this is a solution. |
Ok, I have verified that x=10 is a valid solution, but x=4 does not satisfy this equation. I can tell you that you are wrong. The only issue here is that I cannot determine if x=4 is indeed an extraneous solution or not. I'll just solve the equation.
| \(\sqrt{2x-4}-x+6=0\) | Add everything but the radical expression to the right hand side of the equation. |
| \(\sqrt{2x-4}=x-6\) | Square both sides to eliminate the radical. |
| \(2x-4=(x-6)^2\) | Expand the right hand side of the equation. |
| \(2x-4=x^2-12x+36\) | Move everything to one side of the equation again. |
| \(x^2-14x+40=0\) | This is a quadratic, so there will be another solution to this equation. Factoring is easiest here. |
| \((x-10)(x-4)=0\) | Now, set each factor equal to 0 and solve for x. |
| \(x-10=0 \quad x-4=0\\ \quad\quad\hspace{1mm}x=10\quad\quad\hspace{2mm} x=4\) | |
Ok, you solved the equation correctly, but x=4 is an extraneous solution. However, there is no option choice that states both that x=4 is a solution and x=10 is an extraneous one.
So, would it be safe to say that there is one solution, 10 and 4 is an extraneous solution, or that there are two solutions, 4 and 10? Like, I'm not sure which statement would be the most accurate?
+2440 I understand your confusion. Saying that there is one solution is correct. Saying that x=4 and x=10 are both solutions is incorrect. I would put "There is only one solution: x = 10." I believe this to be the most correct.
For some strange reason, though, it is listed twice in the given options.
