Solving for X in different settings and Sin

[derek916]

As the title suggests, I've have some trouble studying for my algebra exam upcoming in regards to solving for X.

If my writing is unclear, you can view the problem in pdf format here: http://www.csus.edu/math/courses/IADPra ... mFormC.pdf


The first problem (20) I believed I was doing right, but I receive two of the available answers

Root (-x +1) +1 = x

My approach to the problem
A. Moved 1 over to get Root (-x+1) = x-1
B. Squared both sides to clear the root and to get -x +1 = (x-1) (x-1)
C. Moved all to the right side to get x^2 -x
D. Quadratic formula to get 1 (+-) root 1 over 2
E. Answer 1 or 0

Correct answer is C. 1

The second problem (45)

2x^2 +x -3 over x-1 = 5

My approach to the problem
A. Multiplied both sides by x-1 to get 2x^2 +x -3 = 5x -5
B. Cleared to one side to get 2x^2 - 4x + 2 = 0
C. Quadratic formula to get 4 (+-) 16 - 16 over 4
D. Answer 4 (+-) 0 over 4, or 1

The correct answer is no solution (is this because when 1 is plugged back in the denominator is 0, so it's no solution because 0 =5?) Under what circumstances would there be All Real Answers as this was also a possible choice for my answer selection

The last problem perhaps belongs to geometry, but I will post it here since it was included in my algebra exam

This problem requires viewing the triangle drawn in 16 in the pdf

My approach was reading up on SOH CAH TOA
Being SIN i chose the opposite side (I believe is 4) and the Hypotenuse 5

However, the answer was 3/5 which has me confused as I thought the opposite side was opposite of hypotenuse, and adjacent being the short one.

Thanks for viewing and or solving

[maggiemagnet]

The first problem (20) I believed I was doing right, but I receive two of the available answers

Root (-x +1) +1 = x

Did you remember to check your solutions in the original equation?

The second problem (45)

2x^2 +x -3 over x-1 = 5

Did you remember to check your solutions in the original equation?

For radical equations, you can't have negatives inside the (original) square root. For rational equations, you can't have zero in the (original) denominator(s).

Thank you for showing your work so nicely, by the way!

[derek916]

Thanks, yes I did check my question and this was throwing me off...I was wondering why my X never worked out in the equations.



Edit: After reading the 1st solution manual, could I have simply marked no solution as soon as I saw the negative exponent? All of this information is very important as my test will be timed. Thanks.

[maggiemagnet]

Edit: After reading the 1st solution manual, could I have simply marked no solution as soon as I saw the negative exponent?

What "solution manual"? Are you referring to something in the link you gave before? And what negative power? I'm not seeing one in either of your questions.

[derek916]

Sorry , I meant the link. And I now realized thanks to you, that since roots have to be positive I could of just plugged 0 or 1 in for x and finished in 2 minutes