Can You Solve These Rational Equations?

  • Thread starter anonymous12
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In summary, the problem was that the last factor, -4x-4, was added (or subtracted) from the other factors, and so the zeros for the numerator weren't found when looking for the critical points. The answer can be found by solving the Rational Roots test and writing the numerator (x+1)(x+2)(x-1) - 4x - 4) as a cubic with 3 rational roots.
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  • #2
Without actually giving you all the details of the process for #i,

Subtract x from both sides;
Multiply both sides by (x+2);

The left-hand member can be expressed as x2+x-6.
You need to consider two cases. (x+2) positive, and (x+2) negative. Also, as you might have noticed, x cannot be -2.
 
  • #3
anonymous12 said:
k) 4x - 4/ x - 1 < x^2 + 3x + 2
Need parentheses! This looks like
[tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]
Furthermore, looking at your work, it looks like the the numerator was originally 4x + 4. So which is it? 4x - 4 or 4x + 4? Please double check the problem.
 
  • #4
eumyang said:
Need parentheses! This looks like
[tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]
Furthermore, looking at your work, it looks like the the numerator was originally 4x + 4. So which is it? 4x - 4 or 4x + 4? Please double check the problem.

Sorry, the question is (4x+4)/(x-1) < x^2 + 3x + 2 and by the way how did you write the problem as:

[tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]
 
  • #5
anonymous12 said:
Sorry, the question is (4x+4)/(x-1) < x^2 + 3x + 2 and by the way how did you write the problem as:

[tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]

There's a button in the editing window (the last one on the 2nd row, the one that looks like the summation symbol) that allows you to type in LaTeX. You should learn it -- makes the expressions/equations/inequalities much more readable.

Anyway, I think your problem here is that you set up the sign chart too early. You have this:
[tex]\frac{(x+1)(x+2)(x-1)\text{ }(-4x-4)}{x-1} < 0[/tex]
... and the problem is that it looks like the last factor (-4x-4) is multiplied by the other 3 factors, when in fact it is added (or subtracted). And so you didn't get the correct zeros for the numerator when looking for the critical points.

What you'll need to do is take the numerator (which should be written like this):
[tex](x+1)(x+2)(x-1) - 4x - 4[/tex]
multiply out the 3 factors, combine like terms with the -4x - 4, and THEN set equal to zero. Since this will be a cubic, you'll have to try the Rational Roots test. (The 3 roots are all integers, which makes things a little easier.)
 
  • #6
OMG! Thank you so much! This question was driving me crazy and thanks to you I finally solved it! :)
 

1. What is a rational equation?

A rational equation is an equation that contains one or more rational expressions, which are expressions in the form of a fraction where the numerator and denominator are polynomials. These equations can be solved by manipulating the fractions and solving for the variable.

2. How do I solve a rational equation?

To solve a rational equation, you can follow these steps:

  1. Identify any extraneous solutions (values that make the denominator equal to 0).
  2. Find the common denominator of all the fractions in the equation.
  3. Multiply both sides of the equation by the common denominator to eliminate the fractions.
  4. Simplify the resulting equation by combining like terms.
  5. Solve for the variable by isolating it on one side of the equation.
  6. Check your solution by plugging it back into the original equation.

3. Can I cross-multiply to solve a rational equation?

No, cross-multiplying is not a valid method for solving rational equations. It can lead to incorrect solutions and missing potential solutions. It is recommended to follow the proper steps for solving rational equations instead.

4. Can I solve a rational equation graphically?

Yes, rational equations can also be solved graphically. This involves graphing the two expressions on either side of the equation and finding the point(s) of intersection. These points will represent the solutions to the equation.

5. What are extraneous solutions in rational equations?

Extraneous solutions are solutions that may appear to satisfy the equation, but when plugged back into the original equation, make one or more of the denominators equal to 0. These solutions are not valid and should be discarded when solving rational equations.

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