Getting two positive x values when using quadratic

[Nicole Huang 3F]

Hi, so from lecture, Dr. Lavelle talks about how we use the quadratic formula to solve for the x change value in ICE tables, and then proceeds to show how we omit the negative x value and use the positive one. What happens when we get two positive x values? Would we test both x values into the ICE tables or would both of them be correct?

[Sid Panda 3A]

Hi, so from lecture, Dr. Lavelle talks about how we use the quadratic formula to solve for the x change value in ICE tables, and then proceeds to show how we omit the negative x value and use the positive one. What happens when we get two positive x values? Would we test both x values into the ICE tables or would both of them be correct?

yes, you should test both x-values.

Most likely though, one of your x-values would be too big. For example, if you have a ending concentration that is 0.23-x, and your x values are x=0.1,0.5, then only the x=0.1 value would work. If you did 0.23-0.5, you would get a negative concentration for your substance, which is not possible.

If both x-values don't bring up an issue like the one i mentioned above, then I'd test both of them in the equilibrium expression to see if they match with the constant.

[KatarinaReid_3H]

If you arrive at two positive x values then you would have to compare them to the concentration values in the ICE box. One of the x values will likely be larger than the initial concentration and therefore you would arrive at 0 product or reactant wherever you are subtracting the x value. Thus, only one of the x values will make practical sense to use whether the other one is negative or too large for the concentrations you are dealing with.

[Brian Bui 3H]

Test both and one of the values should give you a negative concentration when plugging it into one or more of the equilibrium expressions on an ICE table (ex. 0.2 - x). When this happens you know that that specific value of x can't be used because you can't have a negative concentration! So use the other value of x, which, when plugged in, should give all positive and thus valid concentration values.

[Mahnoor_Wani_1I]

If you want to practice ICE tables with two positive x values, topic 5I has textbook problems that pertains to these types of problems. Usually, one of the positive value, when plugged in, would cause you to have a negative new equilibrium concentration.Therefore, you will be able to eliminate that x value

[Jamie Wang 3C]

Hi, so from lecture, Dr. Lavelle talks about how we use the quadratic formula to solve for the x change value in ICE tables, and then proceeds to show how we omit the negative x value and use the positive one. What happens when we get two positive x values? Would we test both x values into the ICE tables or would both of them be correct?

yes, you should test both x-values.

Most likely though, one of your x-values would be too big. For example, if you have a ending concentration that is 0.23-x, and your x values are x=0.1,0.5, then only the x=0.1 value would work. If you did 0.23-0.5, you would get a negative concentration for your substance, which is not possible.

If both x-values don't bring up an issue like the one i mentioned above, then I'd test both of them in the equilibrium expression to see if they match with the constant.

this was very helpful! I was wondering the same thing. What if the two values would both work? is that an unlikely scenario?

[Chloe Little 3K]

Hi, so from lecture, Dr. Lavelle talks about how we use the quadratic formula to solve for the x change value in ICE tables, and then proceeds to show how we omit the negative x value and use the positive one. What happens when we get two positive x values? Would we test both x values into the ICE tables or would both of them be correct?

yes, you should test both x-values.

Most likely though, one of your x-values would be too big. For example, if you have a ending concentration that is 0.23-x, and your x values are x=0.1,0.5, then only the x=0.1 value would work. If you did 0.23-0.5, you would get a negative concentration for your substance, which is not possible.

If both x-values don't bring up an issue like the one i mentioned above, then I'd test both of them in the equilibrium expression to see if they match with the constant.

this was very helpful! I was wondering the same thing. What if the two values would both work? is that an unlikely scenario?

I think it'd be very unlikely to have two values that work, so I wouldn't worry about it too much.

[Kelly Singh]

I usually just try both and use the most reasonable one

[Kiyoka Kim 3C]

If you get two positive values test them both by plugging them into the equilibrium concentration expression you got from your ICE table. Usually, one should give you a negative value which is not possible, and therefore the other one would be the x value.

[Namita Shyam 3G]

Ya I've also run into this problem a couple times. What you can do is just take both of those values and plug it back into the expressions you made in the "equilibrium" row. If you get a negative value, you know that you can't use that #, but if you get all positive values (because concentrations must always be positive), you're good to go!

[My-Lan Le 1L]

I've run into the same problem, but usually, the correct x value is the smaller number. The larger x value tends to cause the final concentration outcome to be higher than the original concentration, so you would not use it. Nevertheless, I would still plug the larger one back into the concentration equations in the ICE table to be safe, but usually, the smaller number is the correct x value.

[Annette Fishman]

I've run into this problem when completing the Sapling, but even though they are both positive, both shouldn't make sense in the context of the problem. Some people above have noticed that the correct x value is commonly the smaller one. To be sure, though, it is necessary to check for yourself.

[Norah Gidanian 3D]

If you get two positive values it is likely that one of the values was larger than the initial concentration you are given therefore you would omit that one because you cannot end up with a negative concentration.

[Cora Chun 2D]

When you get two positive values after using the quadratic formula, you need to see which one is smaller than the original concentration given. When you plug the x value into the ICE table, all of your equilibrium values should be positive. For example, if you use an x value that is larger than the original concentration given, when it is subtracted from the original value, you will get a negative equilibrium value for that molecule, which is impossible. Hope this helps!

[Frankie Mele 3J]

Typically when you end up having two positive values after solving the quadratic equation, one of the values will be too big or be a number larger than the initial concentration. Since the large number wouldn't make any sense to be in the equation, the other positive value will most likely be the correct x value.

[Sahaj Patel Lec3DisK]

You should either test them, or shift your equation around to maybe make it a bit easier to derive a x-value from. Hope this helps!

[Bella Wachter 1A]

In addition to what everyone else has been saying here, I'd like to add that technically, your x value can be negative. There were several Sapling problems from HW#1 that had negative x value solutions.

What matters, however, is that you never have a negative concentration (because negative concentrations are not possible). So, if you have a positive x-value that produces a negative equilibrium concentration but a negative x-value that produces all positive equilibrium concentrations, the negative x-value would be correct. I know there was at least 1 Sapling problem concerning this (at least with the numbers I was given).

I believe this is only because I didn't flip the equation around when solving (so my products actually decreased and the reactants increased, but I left it in the forward reaction form), but it's good to keep in mind.

[Grecia Velasco 1G]

Hi, so from lecture, Dr. Lavelle talks about how we use the quadratic formula to solve for the x change value in ICE tables, and then proceeds to show how we omit the negative x value and use the positive one. What happens when we get two positive x values? Would we test both x values into the ICE tables or would both of them be correct?

If one of your x values results in a negative concentration when plugged into the eq. concentration equation, then do not use that one. Use the smaller, positive x value in order to have a positive eq. concentration. :>

[Olivia Smith 2E]

If you get two I would just input them into the equation. One of them will most likely be too large and cause a negative number. I would choose the smaller of the two, in the problems we are doing as they are usually with small values. However, I would always do both just in case!

[Vanessa Perez]

if you get two positive x values from the quadratic formula, plug it into your ICE table, if the value outputted then is negative, do not use that x value

[Katie Nye 2F]

This happened to me when I was doing the sapling homework! Most of the time, when it comes to calculating equilibrium concentration, there is an initial value that you subtract x from. The larger value is most likely the one that will not work because once you subtract the larger number from the initial, you get a negative number. That should rule that value out.

[Serena Song 1A]

Try plugging both x values back into the concentration equations from the ICE table and see if they make sense. Usually one of the values will produce a negative concentration value which is impossible so we go with the other one. For example, if the concentration of a compound once it reaches equilibrium is .3 -2x, and the 2 x values you get are .2 and .1, .1 would be the correct x value. This is because .3 - 2 (.2) = -.1M which doesn't make sense logically.

[Joseph Hsing 2C]

With two x values, there's always one value that just doesn't make sense with the numbers you have, like if the value is greater than your inital molar concentrations.

[Madison Muggeo 3H]

Hi! Usually when you get two positive values, one will be greater than the initial concentration you started with, so this value is not usable and you would ignore it. Hope this helps!

[Crystal Pan 2G]

From all the examples we have done so far, when there are two positive values, one of the x values will be larger than the initial concentration we need to subtract x from. Since it's not possible to have a negative concentration, the one that is smaller than the initial concentration will be the answer.

[Karl Yost 1L]

When you have two positive x-values, one will typically not make sense.

For instance, let's consider the reaction XM --> X + M, where you start with a 0.05M solution of XM.

You solve for x and get, e.g., x = 0.0039 and x = 0.5817. We can automatically rule out the x-value of 0.5817, as it is greater than the initial concentration of XM that we have (negative concentrations are impossible).

[Hana Sigsbee 3B]

if both x values are positive one will likely be too big to be an option, like it will be larger than the concentration of one of the reactants or products so in that case, you would use the smaller x value.

[Gwirnowski 3B]

There will always be an x value that is disregarded when you look at the context of the problem. Whether the x value is negative or produces values not possible within the scope of the problem, you will always find it results in one viable x value to use.

[Anusha Chaudhary 1J]

When determining the X-value to use, you can automatically rule out the negative x values, as it is impossible to have a negative concentration. With this idea, we can also ignore any positive x-value that is greater than the initial molar concentrations, as it will give us a negative final concentration.

[Catherine Bubser 2C]

Is it possible that if we used the quadratic formula and got two positive values that both these values would be less than the initial molar concentration?

[Sofia Lombardo 2C]

If both the x values are positive I think we should test the values. There will might likely be a value that, when subtracted from the initial concentration, will give a negative concentration so then it cannot be that value.

[Jaden Joodi 3J]

Yes, you want to test both positive values in order to see which one would fit in the ICE table

[DavidTabib 3H]

You should test both values of x.

[Coleman Morrow]

You should test both of them and in almost all cases one and not the other will work. Then you just have to proceed with the one that works.

[Sai Ramadas 1J]

When you get two positive values, one is likely too big. You can check to see if either of your values are greater than your initial concentration of the reactant. The one that isn't larger than the initial concentration is the correct value.

[AndrewNguyen_2H]

One of the positive values will be larger than one of the initial concentrations for the ICE table which will be impossible.

[Anthony_Sandoval_1D]

You would test both of the values out and the majority of the time only one will work.

[joshtully]

Test both and usually only one will work.

[Aria Movassaghi 1A]

you should test them to see if one might give you a negative concentration. if that is the case, you can rule that value out.

[Jaden Kwon 3C]

When you get 2 positive values usually one of them doesn't work. For example one of your values could be positive but too large which could result in a negative equilibrium concentration for either your reactant or product as you usually have to subtract your calculated value from your initial value which could be negative if that calculated value is too large.

[SashaAnand2J]

Hi Nicole!
Although you can get two positive x values when using the quadratic equation, typically one of the values will be too large and cause a concentration to become a negative (which is not possible). In general, test out both values, but the value that keeps the concentration positive is correct.

[Sonel Raj 3I]

Usually, if you had two positive values, one would be obviously much larger than what you would expect, and you could immediately rule that one out. I would suggest testing both values and seeing whether they match up in the ice table, and whether you concentrations are positive.

[Karen Zheng_2K]

I got the same problem with one of the Sapling problems for this week, I just kind of looked at which number seemed the most feasible and also just plugged it back into the quadratic equation to double check!

[Nicoli Peiris 1B]

When this happens chose the x value that is less than any of the initial concentrations given.