Basic algebra responsible for brain fart

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SUMMARY

The discussion centers on solving the equation 5.10 = [(1+x)/(1-x)]^2 for x in a chemical equilibrium context. The correct solution for x is 0.387, derived through proper algebraic manipulation rather than taking square roots directly. Participants emphasize the importance of considering physical constraints, such as concentration limits, which dictate that x must remain between 0 and 1. Missteps in solving the equation stem from neglecting the implications of negative square roots and the behavior of the function within the specified range.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with algebraic manipulation techniques
  • Knowledge of ICE (Initial, Change, Equilibrium) tables in chemistry
  • Basic understanding of square roots and their implications in equations
NEXT STEPS
  • Study the application of ICE tables in chemical equilibrium problems
  • Learn advanced algebra techniques for solving quadratic equations
  • Explore the implications of physical constraints in chemical calculations
  • Review the properties of square roots and their role in algebraic solutions
USEFUL FOR

Students in chemistry, particularly those tackling equilibrium problems, as well as educators seeking to clarify algebraic methods in scientific contexts.

MurdocJensen
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Homework Statement



I'm trying to solve for x in a chemical equilibrium problem (college chem). Once x is found I have no problem finding equilibrium concentrations. What's important is below.

Homework Equations



5.10 = [(1+x)/(1-x)]^2 --------> x=.387 (right answer)

The Attempt at a Solution



I've tried to take the square root of the right side and cross multiply and solve for x that way, but I keep getting 1. I also tried to expand the numerator with the common denominator to see if I missed something, but solving for x that way eventually got me 1 also. I know there's something simple I'm missing.
 
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Cross multiplying should work. Then you can work out the squares (don't forget that (x + 1)^2 = x^2 + 2x + 1) and bring it to the form
a x^2 + b x + c = 0.
 
That worked awesomely! What I don't understand is why I wasn't able to work it out by taking the square root of both sides. Do you have any insight into that? Not sure if that question's too vague.
 
You can do that, but you may need to consider both positive and negative square root of 5.10,
 
Physical constraints may tell you that x must be positive - and, if x represents a concentration (I can't tell from your original question) then you know 0 < x < 1, and solving by taking square roots works fine, without the need to consider -\sqrt{5.10}
 
That's not true statdad. The equation
\frac{1 + x}{1 - x} = -\sqrt{5.10}
can still give a positive solution for x (there is a minus sign in the denominator) and, in fact, will.
 
CompuChip said:
That's not true statdad. The equation
\frac{1 + x}{1 - x} = -\sqrt{5.10}
can still give a positive solution for x (there is a minus sign in the denominator) and, in fact, will.

Nope: if 0 < x < 1 as I specified, then

<br /> \begin{align*}<br /> 0 &gt; -x &amp; &gt; -1 \\<br /> 1 &gt; 1 - x &amp; &gt; 0<br /> \end{align*}<br />

and no negative term arises.
 
CompuChip said:
That's not true statdad. The equation
\frac{1 + x}{1 - x} = -\sqrt{5.10}
can still give a positive solution for x (there is a minus sign in the denominator) and, in fact, will.
This is correct.
I am guessing this is an equilibrium problem involving an ICE table. If so, then i am going to assume x represents a change in concentration. One of the values obtained for x will not be possible as it will be greater than the initial concentration. Using the positive square-root, the solution should be x=.387. using the negative squareroot, the value for x obtained (if i computed correctly) should be x=2.59. concentrations can be greater than one but if any of the initial concentrations of reagents were smaller than 2.59M, then this value would not be possible, and then the only true solution would be x=.387M (assuming you are working in Molars). MurdocJensen could you post the full question so I could verify this?
Thanks
-Theorem
 

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