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Here are two problems from the High School Mathematics textbook by Hall & Knight. I solved both problems, but when I checked my answers with that at the back of the book, I realised I got only part of the answers correct. I did find the correct solution, so I know I missed a step. What I don't understand is why that step is necessary. I need to understand the concept.

`Higher Algebra' by Hall & Knight

Chapter 2: Proportion, Exercise II, Problems 13 & 14

Problem II-13

Solve the equation:

$$\frac{2x^3-3x^2+x+1}{2x^3-3x^2-x-1}=\frac{3x^3-x^2+5x-13}{3x^3-x^2-5x+13}$$

Solution:

On using Componendo and Dividendo on the given equation we get:

$$\frac{4x^3-6x^2}{2x+2}=\frac{6x^3-2x^2}{10x-26}$$

$$\implies{\frac{x^2(2x-3)}{x+1}=\frac{x^2(3x-1)}{5x-13}}, \tag{A}\label{eqA}$$

[Here I missed a step. I directly cancelled out the $x^2$ on both sides and I reduced the equation to a quadratic and solved for two roots as given below.]

$$\implies{10x^2-15x-26x+39=3x^2-x+3x-1}$$

$$\implies{7x^2-43x+40=0}$$

If $r_1$ and $r_2$ are the two roots of the equation, then $r_1 r_2 = 280$, $r_1 + r_2 = -43$.

$\therefore r_1 = -35, r_2 = -8$

$$\therefore 7x^2-35x-8x+40=0$$

$$\implies 7x(x-5)-8(x-5)=0$$

$$\implies (x-5)(7x-8)=0$$

$$\implies x=5, \ or \ x=\frac{8}{7}$$

[Now, let's go back to $Eq. A$ where I skipped a step. This part I came across elsewhere.]

From $Eq. A$ we get:

$$x^2 \left(\frac{2x-3}{x+1}-\frac{3x-1}{5x-13}\right) =0$$

If $x^2 = 0$, then $x=0$ is a solution.

Now on taking $x \neq 0$ and solving the quadratic we end up with $x=5$ or $x=\frac{8}{7}$

Therefore the solutions for $x$ are:

$$x = 0, \ 5, \ \frac{8}{7}$$

So, I understand that that a cubic equation can have 3 roots (and if less, then they are repeated). But isn't the above given equation a sort of faux-cubic equation, since the common factor $x^2$ can be cancelled out? And if the $x^2$ shouldn't be cancelled out, then what is the rule regarding as to why it can't? Because both methods used (reducing $x^2=0$ and the one I used to cancel out the $x^2$ on both sides) seem reasonable to me. But both can't be correct.

Now here is the next problem in the textbook. It is similar to previous one.

Problem II-14

Solve the equation:

$$\frac{3x^4+x^2-2x-3}{3x^4-x^2+2x+3}=\frac{5x^4+2x^2-7x+3}{5x^4-2x^2+7x-3}$$

Solution:

On using Componendo and Dividendo we get:

$$\frac{6x^4}{2x^2-4x-6}=\frac{10^4}{4x^2-14x+6}, \tag{B} \label{eqB}$$

[Here, too, I missed a step. And after I cancelled out the common factors and multiplied and added the rest I got the following.]

$$\implies 6x^2-21x+9=5x^2-10x-15$$

$$\implies x^2-11x+24=0$$

Now, $r_1 r_2=24$ and $r_1+r_2=-11$. So, $r_1=-8$ and $r_2=-3$. And the quadratic becomes:

$$x^2-3x-8x+24=0$$

$$x(x-3)-8(x-3)=0$$

$$(x-3)(x-8)=0$$

$$\therefore \ x=3, \ or \ x = 8$$

[Now let's go back to $Eq. B$ where I missed a step. This, too, I came across elsewhere and I don't understand the concept.]

From $Eq. B$ we get:

$$x^4 \left(\frac{3}{x^2-2x-3}-\frac{5}{2x^2-7x+3} \right)=0$$

If $x^4 = 0$, then $x = 0$ is a solution.

Now if $x \neq 0$, then we solve the quadratic and get $x=3$, or $x=8$. But that's not the end of it. There is another step where we have to check if these two solutions render the expressions $x^2-2x-3$ and $2x^2-7x+3$ to $0$.

Taking $x=3$ in the above two expressions and they equate to $0$, and on taking $x=8$ they do not equate to $0$.

Therefore $x=3$ is not a solution, while $x=8$ is a solution.

There the answer is:

$$x=0, \ or \ x=8$$

Why did we have to check if those two expressions would equate to $0$? Also, if the highest power of $x$ is $4$, then shouldn't there be four roots? And, of course, why can't we cancel out $x^2$ and $x^4$ as common factors in the above two problems respectively?

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  • $\begingroup$ "Canceling" isn't really a thing. It is shorthand for dividing both sides of the equation by the same thing. However, you can't divide by zero, and so before you can do the step, you have to assume that the thing you are dividing by is non-zero. The logic then branches in two ways: what happens if you can divide, and what happens if you can't. But the equation $xy=xz$ DOES have a solution if $x=0$. $\endgroup$ – Aaron Oct 29 '18 at 23:20
  • $\begingroup$ Just clarifying: So you are saying that we have to check if an expression is rendered to zero, because we do not want zeroes in the denominator? Which is why $x^2$ and $x^4$ can't be "cancelled" out immediately? Is this the same reason as to why (in the second problem) we have to check if the expressions $(x^2-2x-3)$ and $(2x^2-7x+3)$ are not rendered to zero by the solutions $x = 3, \ or \ 8$, because both expressions are located in the denominator? $\endgroup$ – H0tblack Desiat0 Oct 30 '18 at 0:38
  • $\begingroup$ So as a rule-of-thumb, whenever there are variables in the denominator we have to first check if the solution would render the denominator to zero; and if the variable itself is a factor on both sides of the equation, we have to check if it equals zero before we "cancel" it out? $\endgroup$ – H0tblack Desiat0 Oct 30 '18 at 0:39
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    $\begingroup$ Whenever there are variables in the denominator, you have to check to make sure the denominator isn't zero. In addition, whenever you have a factor on both sides of an equation, before you cancel them you need to make sure that factor isn't zero. So $x/x=1$ except when $x=0$, at which point it is undefined. Depending on what you are doing, you can sometimes ignore this (we sometimes say that we have a "removable singularity"), but if you are trying to actually solve equations, weird mistakes can appear if you aren't careful. $\endgroup$ – Aaron Oct 30 '18 at 0:50

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