Where have I gone wrong? quadratic equation

[chemnoob]

I have been taking a chemistry class and have run into this problem in my textbook. I have asked others for help and others have solved the problem for me, but none have explained or shown a step by step process for completing it. If this is in the wrong place, I apologise and please refer me to the correct place.

I would like to know the step by step work to move between these two steps, I believe I understand what to do before and after these steps, but cannot figure out how to calculate this step. Help or referral (videos, etc., but I have already watched many) would be much appreciated.

Thanks in advance.

 

[Deleted member 4993]

I have been taking a chemistry class and have run into this problem in my textbook. I have asked others for help and others have solved the problem for me, but none have explained or shown a step by step process for completing it. If this is in the wrong place, I apologise and please refer me to the correct place.

I would like to know the step by step work to move between these two steps, I believe I understand what to do before and after these steps, but cannot figure out how to calculate this step. Help or referral (videos, etc., but I have already watched many) would be much appreciated.

Thanks in advance.View attachment 21190

You have:

(1.0 * 10^-5)(0.80 - x)(0.20 - x) = 4x²

First - simplify:

(0.80 - x)(0.20 - x)

= 0.80 * (0.20 - x) - x * (0.20 - x)

continue....

If you are still stuck, come back and show your work and point out exactly where you are stuck.

 

[chemnoob]

Thank you for the reply, I have attached my attempt at "3". I understood "1" when I first read it, however, I have been working on "3" for several days and have since forgotten "1". I think "1" will be fine, but still lost on "3", not sure where I am going wrong or if I am way off track.

 

[Deleted member 4993]

Thank you for the reply, I have attached my attempt at "3". I understood "1" when I first read it, however, I have been working on "3" for several days and have since forgotten "1". I think "1" will be fine, but still lost on "3", not sure where I am going wrong or if I am way off track.View attachment 21191

In the line before the last line, you have:

(0.16 * 10^-5 - 1.00 * 10^-5 * x + x²) - 4.0 * 10^-5 * x² = 0

then next step should be - take the parentheses out and collect the terms with x² :

0.16 * 10^-5 - 1.00 * 10^-5 * x + (1- 4.0 * 10^-5) * x² = 0

continue,,,,

 

[chemnoob]

I thought I saw where I went wrong, but got lost at where the (1-4.0*10^5)*x^2 came in...?
Thank you again, sorry I'm not fully understanding

 

[Steven G]

I have been taking a chemistry class and have run into this problem in my textbook. I have asked others for help and others have solved the problem for me, but none have explained or shown a step by step process for completing it. If this is in the wrong place, I apologise and please refer me to the correct place.

I would like to know the step by step work to move between these two steps, I believe I understand what to do before and after these steps, but cannot figure out how to calculate this step. Help or referral (videos, etc., but I have already watched many) would be much appreciated.

Thanks in advance.View attachment 21190

As you did in one of your post you can ignore the leading 1.0.
10^-5( .16 - 1x + 1x^2) = 4x^2
.16*10^-5 - 10^-5 x+ 10^-5 x^2 = 4x^2
4x^2 - 10^-5 x^2 + 10^-5x - .16*10^-5 = 0
(4 - 10^-5)x^2 + 10^-5 x - .16*10^-5 = 0 or (4 - 1*10^-5)x^2 + 1*10^-5 x - .16*10^-5 = 0

 

[chemnoob]

As you did in one of your post you can ignore the leading 1.0.
10^-5( .16 - 1x + 1x^2) = 4x^2
.16*10^-5 - 10^-5 x+ 10^-5 x^2 = 4x^2
4x^2 - 10^-5 x^2 + 10^-5x - .16*10^-5 = 0
(4 - 10^-5)x^2 + 10^-5 x - .16*10^-5 = 0 or (4 - 1*10^-5)x^2 + 1*10^-5 x - .16*10^-5 = 0

that is excellent, thank you very much for your help!!