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I have an equation:

$$\left(\frac{b}{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0$$

The question is by factorizing what are the solutions for a?

I am not sure how to do this: I have reduced the equation to:

$$\frac{b^2}{x^3}+x+a=0\\ x^4 - b^2 + ax^3 = 0 $$

An example question (which is not related to the above), is very difficult to understand and only gives one real solution, but may be useful in those trying to help:

The question states if:

$$x^3+xb+a=0 \\$$

one of the solutions is: $$a=2\cdot \left(\frac b3\right)^{3/2}$$

How did they arrive at this?

John
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  • what is $x$ here? Usually it is an unknown. What is it here? – Guy Mar 22 '14 at 15:57
  • thanks for the question, x,b and a are all unknowns – John Mar 22 '14 at 16:05
  • The $b^2$ term needs to be subtracted. When you multiplied out the factors, you should have gotten $$\left(\frac b{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0 \iff \frac bx -\frac{b^2}{x^3} + x - \frac bx + a = 0$$ – amWhy Mar 22 '14 at 16:14

1 Answers1

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We have the equation: $$\left(\frac b{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0$$

and hence the domain of $x$ must exclude $x = 0$.

$$\begin{align} \left(\frac b{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0 & \iff \frac bx -\frac{b^2}{x^3} + x - \frac bx + a = 0 \\ \\ &\iff x-\frac{b^2}{x^3} +a =0\\ \\& \iff \frac{x^4 - b^2+ ax^3}{x^3}=0\end{align}$$

$$x^4 - b^2 + ax^3 = 0 \iff x - \frac{b^2}{x^3} + a = 0 \implies a = \frac{b^2}{x^3} - x$$

Please note that you erred when you went from $$x^4 + b^2 + ax^3 = 0 $$

to $$x^3 + bx + a = 0$$ These are not the same equations.

Regarding your second question, to determine whether the given $a$ is a solution, plug that solution into $x^3 + bx + a = 0$.

amWhy
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  • Will you please stop rewriting your question, based on my answer, and refrain from asking additional questions after already having asked one. For those of us who want to help, we end up chasing a "moving target", and without acknowledging edits, it can leave correct answers seem to not address the (modified) question. – amWhy Mar 22 '14 at 16:21
  • sincere apologies, I'm new to this kind of forum and mathematics, I'll refrain from now on. – John Mar 22 '14 at 16:25
  • Welcome, John! Not a problem. You'll get the hang of this site in no time. (Also: We certainly appreciate your efforts to format you post!) – amWhy Mar 22 '14 at 16:31
  • If I notice any errors in the future, should I add a comment underneath? – John Mar 22 '14 at 16:32
  • Sure, but you can also edit in your post, immediately below the original equation/ formula, etc. e.g. [original equation] followed by: $$\text{EDIT: I'm correcting an error. It should be } \cdots$$ – amWhy Mar 22 '14 at 16:34
  • Thank you very much. If: $$a=(b^2/x^3)-x \$$ is there a way of expressing "a" in terms of just "b" and not include "x"? – John Mar 22 '14 at 17:01
  • No, there isn't. Any solution to $a$ depends on both b and $x$. – amWhy Mar 22 '14 at 17:04
  • Thank you very much amWhy. Could I ask if I wanted to ask another question related to this, i.e. what are the solutions for x, should I start another thread? – John Mar 22 '14 at 18:35
  • Brilliant thank you! :) – John Mar 24 '14 at 11:12
  • You're welcome, John! – amWhy Mar 24 '14 at 11:21