SOLUTION: Hi, I have done this problem in solving equation: 2^2 � x � 1 = 0 This is how I did the problem Factor the equation: (2x + 1) (x � 1) = 0 Factor separately to equal 0: 2x + 1=

Question 141026: Hi, I have done this problem in solving equation:
2^2 � x � 1 = 0
This is how I did the problem
Factor the equation: (2x + 1) (x � 1) = 0
Factor separately to equal 0: 2x + 1= 0 x � 1 = 0
2x = -1 x = 1
x = - �
My question is how to check the answer, and to go back to original equation?
I hope some one can help me please I am not sure about how to go back to the original equation. Thank you very much.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2x^2 � x � 1 = 0
2x^2-2x+x-1 = 0
2x(x-1)+(x-1) = 0
(x-1)(2x+1) = 0
x-1 = 0 or 2x+1 = 0
x = 1 or x = -1/2
----------------------
Checking:
x=1:
2*1^2 - 1 -1 = 0
2-1-1 = 0
0=0
----------
x=-1/2
2(-1/2)^2-(-1/2)-1 = 0
2(1/4) + 1/2 - 1 = 0
1/2 + 1/2 - 1 = 0
1 - 1 = 0
0 = 0
===============
Cheers,
Stan H.

RELATED QUESTIONS
Please help me with the following problem Solve the equation 1/2x2 -4x +8 = 0 this (answered by stanbon)

Hi- I have been working on this problem and I think I am doing it right. I am about... (answered by jim_thompson5910)

I am helping my son with SAT test taking skills. I have done what I can to look at the... (answered by stanbon)

Dear Sir/Madam, I am confronted with the following problem: "Find a polar equation... (answered by ichudov)

my problem is solving for y in the point slope equation. i find the slope with no... (answered by rapaljer)

How does this work, what is this called? If I have a problem like: {{{ 2x^2+5x-3=0 }}}... (answered by ankor@dixie-net.com,Alan3354)

I really need your help in solving this problem. I appreciate your time and efforts. Have (answered by josgarithmetic)

Please help me solve this problem: y^4/3 = -3y This is what I have done: y^4/3 +... (answered by stanbon,ankor@dixie-net.com)

solve by factoring check each answer 2x^2+3x=2 2x^2+3x-2=0 (2x-1)(x+2)=0 x=1/2... (answered by ankor@dixie-net.com)