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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?

Guest Nov 29, 2017
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6+0 Answers

 #1
avatar+79846 
+1

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  !!!

 

 

cool cool cool

CPhill  Nov 29, 2017
 #2
avatar+1493 
+1

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.

TheXSquaredFactor  Nov 29, 2017
 #3
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+1

I mean the (B) equation, thanks for the help

Guest Nov 30, 2017
 #4
avatar+1493 
+1

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.

TheXSquaredFactor  Nov 30, 2017
 #5
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0

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?

Guest Nov 30, 2017
edited by Guest  Nov 30, 2017
 #6
avatar+1493 
0

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.

TheXSquaredFactor  Dec 1, 2017

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