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

#1**+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 !!!

CPhill Nov 29, 2017

#2**+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

#4**+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**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**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