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$$\frac{3x^3 - x^2}{ 9x^2 - 6x + 1 }= x^2$$

I'm trying to solve this problem for some time now, and the answer I received was x=3 Which I realized was a mistake once I checked.

My order of actions was the following:

1) I used the quadratic formula on the denominator (9x^2 etc..) and got 1/3.

2) I subtracted the numbers in the nominator: $3x^3 - x^2 = 3x$

3) I Tripled the nominator in order to reach a common denominator 3x / 1/3 * 3 = x^2 / 1

4) $9x = x^2$

5) $x = 3$

I am not looking for the right answer, I really rather struggle and keep trying new ways in order to reach a solution by myself, but I have hit a wall, and despite the answer being wrong, I am unable to see anything faulty in my way of solving the problem, some guidance would be of much help! I have some doubts about the quadratic formula being used here, but then, why wouldn't it be used here? technically, I should be able to utilize this formula in any situation when met with this format, shouldn't I? I'm trying to fully understand the material yet I keep lagging behind, any help would be greatly appreciated!

I apologize in advance for not formatting the math correctly, I just have no idea how to go about it.

Thanks everyone!

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    $\begingroup$ "I subtracted the numbers in the nominator: 3x^3 - x^2 = 3x 3)"... $3x^3 - x^2$ can not be combined, instead factor out a common term and you you should be able to cancel out a term with your factored denominator (which by the way is not going to be $1 \over 3$). $\endgroup$ – CSCFCEM Aug 7 '15 at 12:39
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    $\begingroup$ It's like almost every step of yours is faulty. Namely: $1,\,2,\,4,\,5$. Of course, just one faulty step is enough. $\endgroup$ – dbanet Aug 7 '15 at 13:06
  • $\begingroup$ But why Can't I use a quadratic formula on the denominator? I get that the answer does not make sense, but why is the first step wrong? $\endgroup$ – Serj Rotaru Aug 7 '15 at 13:42
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    $\begingroup$ @SerjRotaru Using the quadratic formula allows you to write the denominator as $9(x-1/3)^2$ (since this is equal to $9x^2-6x+1$ if you expand it out). It is not telling you that $9x^2-6x+1 = 1/3$. $\endgroup$ – Casteels Aug 7 '15 at 14:10
  • $\begingroup$ You can't use the quadratic formula directly in the denominator because the denominator is not an equation, its an expression. You can use it to factor though. $$\frac{P(x)}{ax^2+bx+c}=\frac{P(x)}{a\bigg(x+\frac{b+\sqrt{b^2-4ac}}{2a}\bigg) \bigg(x+\frac{b-\sqrt{b^2-4ac}}{2a}\bigg)}$$ Or in your case $$\frac{3x^3-x^2}{9x^2-6x+1}=\frac{3x^3-x^2}{9(x+\frac{-6+\sqrt{(-6)^2-4 \cdot 9 \cdot 1}}{2\cdot 9})(x+\frac{-6-\sqrt{(-6)^2-4\cdot 9\cdot1}}{2\cdot 9})}=\frac{3x^3-x^2}{9(x-\frac{1}{3})^2}=\frac{3x^3-x^2}{(3x-1)^2}$$ $\endgroup$ – John Joy Aug 7 '15 at 14:16
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$$\frac{3x^3-x^2}{9x^2-6x+1}=x^2$$ $$\frac{x^2(3x-1)}{(3x-1)^2}=x^2$$ $$\dots$$

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Note that $$ 0=\frac{3x^3 - x^2 }{9x^2 - 6x + 1}-x^2=\frac{x^2( 2- 3x)}{3x - 1},$$ because $9x^2 - 6x + 1=(3x-1)^2$. Hence $x=\frac{2}{3}$ or $x=0$.

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Notice, $$\frac{3x^3-x^2}{9x^2-6x+1}=x^2$$ $$\frac{3x^3-x^2}{9x^2-6x+1}=x^2$$ $$9x^3-9x^2+2x=0$$ $$x(9x^2-9x+2)=0$$ $$x(3x-2)(3x-1)=0$$ $$\implies x=0, \frac{1}{3}, \frac{2}{3}$$ Now, substituting the value $x=\frac{1}{3}$ we get undefined value of form $\frac{0}{0}$ on LHS

Hence, by discarding $x=\frac{1}{3}$, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=0, \frac{2}{3}}}$$

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  • $\begingroup$ Thank you for checking my answer $\endgroup$ – Harish Chandra Rajpoot Aug 7 '15 at 12:56
  • $\begingroup$ -1 Zero times what number equals zero? The answer is clearly $(\frac{1}{3})^2$ $\endgroup$ – John Joy Aug 7 '15 at 14:27

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