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Consider the equation $x^2 - 5x + 6 = 0$. By factorising I get $(x-3)(x-2) = 0$. Which means it represents a pair of straight lines, namely $x-2 =0 $ and $x- 3 = 0$, but when I plot $x^2 - 5x + 6 = 0$, I get a parabola, not a pair of straight lines. Why?

Plotting: x^2 - 5x + 6 = 0

at Wolfram Alpha, I get the result:

enter image description here

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    @lamar Thanks for the edit. I really should have included the graph earlier, stupid me. – 2017-02-18
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    As the answers explain, the "red dots" on the plot at 2 and 3 are the solutions. The parabola you see is $f(x) = x^2 - 5x + 6$ and has been kindly thrown in for reference. You don't see "a pair of lines" on the plot because those lines are being multiplied together (making a parabola). If you also plot $g(x) = x - 2$ and $h(x) = x - 3$ you will observe two parallel lines that cross the x axis (have roots / intercepts) at different points (2 and 3 respectively). – 2018-10-25

8 Answers 8

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$x^2-5x+6=0\quad$ is not the equation of the parabola.

The equation of the parabola is $\quad x^2-5x+6=y(x)$

$(x-3)(x-2)=0\quad$ is not the equation of two straight lines.

The equations of the two straight lines is $\quad x-3=y(x)\quad\text{and}\quad x-6=y(x)\quad$ or : $$(x-2-y)(x-3-y)=0$$

Do not confuse the "equation" of a curve with the "equation" to be solved for an unknown $x$.

The meaning of the word "equation" isn't the same. In the first case, it means a relationship between two variables $y$ and $x$. In the second case, it means an equality not for any values of $x$, but only for some particular values of $x$. Then, solving for $x$ means finding those particular values.

In addition :

In the very different case $\quad x^2-4xy-y^2=0\quad$ there is $y$ in the equation. This is a relationship between $y$ and $x$. So, it is valid for various values of $x$ and the related values of $y$. This allows to draw a curve $$y(x)=(2\pm \sqrt{5})\:x$$ So, two straight lines : $\quad y(x)=(2+ \sqrt{5})\:x \quad$ and $\quad y(x)=(2- \sqrt{5})\:x$

Further addition :

$x^2-5x+6=0\quad$ is commonly understood as to be solve for $x$, that is, to find the roots of the equation. The answer is two constant values : $x=2$ and $x=3$.

If one want to make understand that the question is not to find the roots of the equation on the common sens, but is to find some unknown relationship between $x$ and $y$ satisfying $x^2-5x+6=0$ , in order to avoid the ambiguity, the equation should be written as : $$\left( x(y)\right)^2-5x(y)+6=0$$ because, this specifies that $y$ exists and that the equation have to be solved for a function $x(y)$.

Solving it leads to $$x(y)-2=0 \quad\to\quad x(y)=2$$ $$x(y)-3=0 \quad\to\quad x(y)=3$$ that is two lines parallel to the y-axis.

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    Just out of curiosity, how would you describe the graph of $y^2=x^2$? – 2017-02-17
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    $y^2=x^2$ is a relationship between two variables $y$ and $x$. So it is the equation of a curve which is made of two straight lines : $y(x)=\pm x$ , or $\left( y(x)=x \text{ and } y(x)=-x \right)$. – 2017-02-17
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    @JJacquelin: There is an ambiguity, things are not as straightforward: $x^2-5x+6=0$ represents a curve if it is regarded as the set of points in the plane: $\{ (x,y)\mid x^2-5x+6=0 \}$ – 2017-02-17
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    @JJacquelin I entered https://www.wolframalpha.com/input/?i=x%5E2+%2B+4xy+%2B+y%5E2+%3D+0 and I got a pair of striaght lines. This conflicts with third line of your answer. – 2017-02-17
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    @ P Vanchinathan : in $\quad x^2-5x+6=0\quad$ where is no $y$ at all. So, from where this $y$ in your $(x,y)$ is coming ? Of course, if you modify the sens of the equation $x^2-5x+6=0$ in implicitly putting it into a different field of mathematics where two isolated points $(2,0)$ and $(6,0)$ are called a "curve", one can discuss about it. As one can discuss about "how many angels can dance on the head of a pin" . – 2017-02-17
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    Actually the meaning of the word "equation" *is* the same. The meaning is that it is a statement of equality (of the RHS and LHS). The meaning (of the word) doesn't change meaning because you happen to put in one or two variables in the statement. – 2017-02-17
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    @ skyking : Of course, the symbol = means the equality of the RHS and the LHS in both cases. The point isn't here. The point is what are the meanings of the LHS and the RHS. Do you argue that the meaning of $=0$ and $=y(x)$ is the same ? – 2017-02-17
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    Just because the equation has only 1 variable, it doesn't mean it can't be plotted in 2. $x^2+y^2=1$ is a circle in 2 dimensions and a cylinder in 3. The given equation can be 2 points on a number line or, if plotted in 2 dimensions, 2 lines. – 2017-02-17
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    @Mike : Of course, I agree. But if you set an equation such as $x^2-5x+6=0$ to be solve for $x$, there is no $y$, no $z$, no other variable in it. As I already said, if afterwards one add some specifications (for example: "in 2 dimensions" or "in three dimensions") of course the meaning is not the same as is was in the original wording. – 2017-02-17
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    Sorry for this general remark : For me, all this stuff is silly : Simply, the original question comes from the confusion between an equation to be solved for a variable $x$ and an equation of curve $y=f(x)$. Why taking pleasure in making things more complicated that they are ? – 2017-02-17
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    @JJacquelin Did you see my comment under your answer ? What would say for $x^2 - 4xy - y^2 =0 $ ? The plot gives two lines on Wolfram alpha. – 2017-02-17
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    @ A---B : In your original question the equation is $x^2-5x+6=0$. My answer is about the original question, not about a modified equation $x^2-4xy-y^2=0$ which is very different. See the addition to my first answer. – 2017-02-17
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    Wrong. In 2-D Cartesian coordinates system $(x−3)(x−2)=0$ actually **is** an equation of two lines, namely two lines parallel to $OY$ axis, one line $x=2$ and another one $x=3$. – 2017-02-17
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    @ CiaPan : No, $\quad (x-3)(x-2)=0\quad$ isn't an equation of curve. The equation of curve is $\quad(x(y)-3)(x(y)-2)=0\quad$. Using $x$ instead $x(y)$ is a shorthand which confuses a number with a function. Of course, it can be used commonly and it is commonly used. But in a discussion on this subject, one have to be exact on the words and symbols, without implicit meaning. – 2017-02-17
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    @JJacquelin Can you show me the plot of $x(y)^2 - 5x(y) + 6 = 0$ ? please. – 2017-02-17
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    @ A---B : Yes I can, but you can do it as well : Draw two vertical lines (parallel to the y-axis), one with abscissa$=2$, the other with abscissa$=3$. – 2017-02-17
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    @JJacquelin On wolfram alpha I mean. – 2017-02-17
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    Do you realy need WolframAlpha to draw a so simple function ? . If you cannot, contact the WolframAlpha team. I am not associated with WolframAlpha. – 2017-02-17
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    @JJacquelin No you are misunderstanding. This is not my homework, I have not being told to graph this function, I just wanted to know if wolfram alpha can plot a function like this. – 2017-02-17
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    @JJacquelin Thanks for the answer and for amending as per my request. – 2017-02-17
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    It may also be helpful to mention that $x^2-5x+6=0$ can also be thought of in terms of finding the intersection between the curves described by $y(x) = x^2-5x+6$ and $y(x) = 0$. +1 – 2017-02-17
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    @P Vanchinathan $ x^2-5x+6=0 $ does not represent any curve.. Simply saying (x,y) without connecting them together does not amount to any curve definition. – 2017-02-18
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    @ A---B : Sorry for the misunderstanding. Now, about WolframAlpha : The request $x^2-5x+6=0$ is interpreted as solving the equation for $x$. Apparently, to make it interpreting as a request to draw $y$ as a function of $x$, the equation must include $y$ in it. For example, try : http://m.wolframalpha.com/input/?i=plot+%28x%5E2-5x%2B6%29y%3D0&x=0&y=0 – 2017-02-18
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    @JJacquelin So, you're saying $(x−3)(x−2)=0$ is not an equation of a curve? Will you dare to say also $x^2+y^2=16$ is not an equation of a curve? AFAIK it is an [equation of a circle](https://en.wikipedia.org/wiki/Equation_of_a_circle), and a circle _is_ a curve... – 2017-02-18
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    @JJacquelin You're absolutely wrong. Using $x$ in equations is **not** any _'shorthand for $x(y)$'_ or for anything else. Symbol $x$ is just a name of a variable, whilst $x(y)$ denotes a function $x$ with argument $y$. And we do **not** use $x(y)$ in equations, because in many cases we do not know in advance, whether the equation resolves to a functional form $x(y)$ or $y(x)$ ...or **none**, as in the aforementioned case of a circle. (...) – 2017-02-18
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    @JJacquelin (contd.) For another example, $9x^2 - 12xy + 4y^2 - 2x - 3y + 7 = 0$ is an equation of a parabola, but the equation can't be transformed to the form of $y(x)$ or $x(y)$ function (see it [plotted in Wolfram Alpha](https://www.wolframalpha.com/input/?i=plot+9x%5E2+-+12xy+%2B+4y%5E2+-+2x+-+3y+%2B+7+%3D+0)). – 2017-02-18
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    @CiaPan : You are absolutely wrong. $(x-3)(x-2)=0$ is not an equation of a curve in the strict sens of equation of function. $(x(y)-3)(x(y)-2)=0$ is an equation of a curve. $x^2+y^2=16$ is an equation of a curve. Sorry, I will no longer answer to people who post without reading all what was discussed before. – 2017-02-18
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This is because Wolframalpha is plotting $y=(x-2)(x-3)$, which is a parabola.

As you have entered $(x-2)(x-3)=0$, it is merely indicating where the intersection is between $y=(x-2)(x-3)$ and $y=0$, which is why there are red dots on the points where the $x$-coordinates are $2$ and $3$.

This is not interpreted as a function by Wolframalpha as it contains only one variable, and thus it is commonly interpreted as mentioned above, but given that $x $ is some function of $y $, it will be, as you have suggested, two lines parallel to the $y $-axis.

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    But when I enter https://www.wolframalpha.com/input/?i=x%5E2+%2B+4xy+%2B+y%5E2+%3D+0 I do get a pair of striaght lines. – 2017-02-17
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    Because you provided independent variable $x$ and $y$ and not the same $x$ twice. – 2017-02-17
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    @A---B Also look further down for the solutions for $y$ as function of $x$ which you get by solving the quadratic (you could solve for x too and get two solutions too). The two solutions describe a line each. – 2017-02-17
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    @A---B Alternatively you could complete the square as shown $x^2+4xy+y^2 = (x+2y)^2-3y^2 = (x+2y-\sqrt3 y)(x+2y+\sqrt3 y)$. That expression is zero only when one of the factors are, that is $x+2y-\sqrt3 y=0$ or $x+2y+\sqrt3 y=0$ (which are equations of lines). – 2017-02-17
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    @mvw I can write $(1 \cdot x - 0 \cdot y - 2)(1\cdot x - 0\cdot y - 3) = 0$, what then ? – 2017-02-17
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    @skyking yes so does the equation $(x-2)(x-3) = 0$ but I get a parabola for plotting this equation. – 2017-02-17
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    @A---B Yes, but if you study the https://www.wolframalpha.com/input/?i=(x-2)(x-3)%3D0 carefully you see that in the former it says "implicit plot" and in the later "root plot" - they relate quite differently to the equation. The "implicit plot" is the plot of all solutions to the equation (compare the equation for a line where the line is the implicit plot), while the "root plot" is the plot of the LHS and the solutions are where the curve intersects the x-axis (marked with red dots). – 2017-02-17
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    @A---B If you write an equation in one variable with non-zero RHS both sides will be plotted (and the plot will contain a legend describing what has been plot). See for example https://www.wolframalpha.com/input/?i=(x-2)(x-3)%3D4x where again the solutions to the equation are marked with red dots (well the solution is the $x$ value at those points - the $y$ value is the value the LHS and RHS side take for the solution). – 2017-02-17
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After much discussion in comments, I have decided to interpret this as a WolframAlpha question. Many people would not plot an equation in one variable. The solution could be plotted on a number line. In $2$ dimensions, the plot would in fact be $2$ lines parallel to the $y$-axis.

WolframAlpha does not interpret it in this fashion. It seems to interpret each side of the equation as a function in $1$ variable. $f(x)=x^2-5x+6,g(x)=0$. It graphs both and highlights points of intersection.

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    If the question is interpreted as a WolframAlpha question, your answer is pertinent. I would only add this : To make WolframAlpha understand $x^2-5x+6=0$ is not to be solved for $x$, one have to specify that $y$ exists, that is to say to look for a relationship between $x$ and $y$. The equation must be written as : x(y)^2-5*x(y)+6=0 . With this specification $x(y)$ instead of $x$, WolframAlpha will answer that the solutions are two parallel lines. – 2017-02-17
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I believe the other answers are correct in the interpretation of the question and in the way that they have addressed the issue. But I would like to look at the question a bit differently. Whether this is of any use or interest is up to the OP.

$y = (x -2)$ and $y = (x-3)$ are lines. So why is their product a parabola?

Consider $y = (x-2)(x -3) = (x-2)x - 3(x-2)$. If you compare this to the slope-intercept form of a line, $y = mx + b$, then you get $$m = (x-2)\\b=-3(x-2)$$

In particular, it is the $m = (x-2)$ part that makes this a parabola instead of a line. In a line, the slope is constant. But in this parabola, the slope is changing in a linear fashion itself. This is not just true of this parabola, but of any parabola. That is, a parabola is just a "line" whose slope changes linearly as you go along - a relation that is more explicit when examined using calculus.

So the parabola is indeed a combination of two lines. Only the combination is by one of the lines modifying the slope of the other "line".

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    Ok, but consider "x^2 -4xy + y^2 = 0$ which actually when plotted on WA gives a pair of lines, What do you think ? – 2017-02-17
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    $x^2 -4xy +y^2 = 0$ is not the equation of a parabola, which (when oriented with the coordinate axes) will mix 2nd order terms of one variable with only 1st order terms in the other ($x^2$ or $y^2$ will be present, but not both). All equations of the form $ax^2 + bxy + cy^2 + dx + ey +f = 0$ are called conic sections, because the set of all $(x,y)$ that satisfy them can be expressed as intersections of a plane and a double-cone For this equation, the plane passes through the cone vertex and contains the cone axis, so the intersection is two lines. – 2017-02-17
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    A side question. Is there any use of that conic formula ? It seems like a big formula and I think it will make thing unnecessarily complicated. – 2017-02-18
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    The use of the formula is to recognize the equation of a conic section when you see it. Because I am familiar with it, I knew at a glance that your equation was one. And because I know what forms are possible for conic sections (hyperbola, parabola, ellipse (includes circle), two lines, one line, a single point), I understood why the graph was two lines immediately. There are also methods for identifying the section from the equation (easy when $b = 0$, but harder otherwise) and transforming the equation to more useful forms, but they are rarely needed, so I don't remember them. – 2017-02-18
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I get $(x-3)(x-2) = 0$. which means it represent a pair of striaght lines namely $x-2 =0 $ and $x- 3 = 0$

When we just write $$ (x-3)(x-2) = 0 $$ and ask for $x$ we usually mean the set of solutions $$ S = \{ x \mid (x-3)(x-2) = 0 \} $$

Two vertical lines in two dimensions we can model as: $$ L_1 = \{ (x,y) \mid (x-3) = 0 \} \\ L_2 = \{ (x,y) \mid (x-2) = 0 \} $$ These are different sets. We sloppily write $x-3=0$ and $x-2=0$ but mean the above.

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    @S.C.B. Why wouldn't it be downvoted? What part of this answer is correct or useful? $(x-3)(x-2)=0$ implies either $x=3$ or $x=2$. It doesn't matter if each factor is constant. This is the entire basis on how we solve quadratic equations. And if solutions in the form $(x,y)$ are plotted on a graph, it makes sense the graph would involve vertical lines since $y$ is absent from the equation. – 2017-02-17
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    As I already said, it doesn't matter if either factor is constant. $y=(x-3)(x-2)$ is a parabola. $(x-3)(x-2)=0$ is in fact 2 vertical lines. Give one point that satisfies this formula where $x$ is equal to neither $2$ nor $3$. – 2017-02-17
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    I made the interpretations explicit. – 2017-02-17
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    And how would you describe $L_3=\{ (x,y)\mid (x-3)(x-2)=0\}$? Is it not the union of $L_1$ and $L_2$? And are you answering the question that is asked? – 2017-02-17
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    @mvw But can you tell why WA differentiate between $x^2-5x+6 =0$ and $x^2 -4xy +y^2 = 0$ ? – 2017-02-17
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    @Mike I thought so. – 2017-02-17
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    @A---B In the first case WA assumes you want $\{ x \mid x^2 -5x+6 = 0 \}$ and in the second it assumes you want $\{(x,y) \mid x^2-4xy+y^2=0\}$. – 2017-02-17
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To get from the equation $$x^2 - 5x + 6 = 0$$ to a set of solutions, one must do some interpretation. Namely, in what universe or "base set" do we search the solutions?

If this is to be interpreted as $\{ (x,y)\in\mathbb{R}^2 \mid x^2 - 5x + 6 = 0 \}$, then the solution set is indeed two vertical lines: $$\{ (x,y)\in\mathbb{R}^2 \mid x=2 \quad\mathrm{or}\quad x=3 \} = \{ (x,y)\in\mathbb{R}^2 \mid x=2 \} \cup \{ (x,y)\in\mathbb{R}^2 \mid x=3 \}$$

However, one could also interprete as the set $\{ x\in\mathbb{R} \mid x^2 - 5x + 6 = 0 \}$ and then your solution set is: $$\{ x\in\mathbb{R} \mid x=2 \quad\mathrm{or}\quad x=3 \} = \{ 2,3 \}$$

Wolfram Alpha took the last approach and showed with two red dots on the $x$ axis (abscissa) the two-point set. Then, it also plotted the difference between the left-hand side and the right-hand side of the equation $x^2 - 5x + 6 = 0$ with another color, along a vertical coordinate axis (ordinate). In a sense, this shows how "far" from being a solution other $x$ values (than $2$ or $3$) are.

In fact, the blue "helpful" addition to the two-point solution set, is the same as $y = (x^2 - 5x + 6) - 0$ or just $y=x^2 - 5x + 6$.

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    In the same spirit, if we search for solutions in $\mathbb{R}^3$, i.e. formally have $$\{ (x,y,z)\in\mathbb{R}^3 \mid x^2 - 5x + 6 = 0 \}$$ then the solution set consists of two parallel _planes_. – 2017-02-17
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As has been pointed out $x^2-5x+6=0\quad$ is not the equation of the parabola.

The equation of the parabola is:

$y_p = x^2-5x+6 = (x-2)(x-3)$

The equations of the two straight lines within the equation are is $y_1 = x-2$ and $ y_2 = x - 3$ and both lines can be plotted as a function of the same variable $x$. But if we multiple the two lines together we get:

$y_1y_2 = (x-2)(x-3)$

and $y_2= y_1 - 1$ so

$y_1(y_1-1) = (x-2)(x-3)$
$y_1^2 -y_1 = (x-2)(x-3)$

and thus $y_p = y_1^2 - y_1$ which isn't terribly useful.

But as $x$ goes to infinity the constant term 6 in the equation becomes negligible. Thus the parabola approaches the curve $y_3 = x^2-5x$ as $x$ approaches $+\infty$ or as $x$ approaches $-\infty$.

Now if we translate the axis to the point ($\dfrac{5}{2}$, $-\dfrac{1}{4}$), we can define:

$y' = y + \dfrac{1}{4}$ and $x' = x - \dfrac{5}{2}$

so that

$y' = x'^2$

and the quandary would seem to disappear.

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You're asking about two different things in your question and in its title.

As for the title:

Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola?

the answer is: it doesn't.

The equation $x^2 - 5x + 6 = 0$ is just an equation, and it does not represent anything unless you define the way of interpreting.
For example, if you say 'it's an equation of one real variable $x$' then it will mean just that: equality of two algebraic expressions, which holds for two real values only: for $x=2$ or $x=3$.
But if you say 'it's an equation of a figure in 3D spherical coordinates, with $x$ denoting a distance from origin', the equation will represent two concentric spherical surfaces.
And for 'it's an equation of a figure in planar Cartesian coordinates XY' it will represent two parallel lines.

Any way it won't represent a parabola.

As for the question itself:

when I plot $x^2 - 5x + 6 = 0$ (at Wolfram Alpha), I get a parabola, not a pair of straight lines. Why?

The shortest answer is: because you plot it at Wolfram Alpha.

The site performs plenty of pre-processing of users' queries, so that it can usually make a mathematical sense of questions asked in nearly everyday's prose. That applies also to equations. And your equation has a form $f(x)=0$ which often appears in questions like "what are zero points of a function $f(x)$?" So WAlpha tries to explain the problem to you by plotting $y=f(x)$ (which is a parabola!) and marking its zero points.
But the parabola here is not the 'answer' – it is just an illustration of the nature of the problem (as Wolfram Alpha 'thinks' you have). The actual answer is just a pair of red dots on the OX axis.

You can make this way of iterpreting invalid to WAlpha by adding explicit $y$ term, which renders any attempt to extract $y(x)$ doomed. That adds another line to your plot, but allows you to see the two lines you expected:

plot y(x^2 - 5x +6) = 0

results in this plot:

enter image description here

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    The example of Plot y(x^5-5x+6)=0 in WolftanAlpha has been cited in comments a long time ago. – 2017-02-18
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    @JJacquelin I have read every single comment but I can't find it. Can you please show me where you have seen it ? – 2017-02-18
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    This was in a comment addressed to you. Copy of the comment : @ A---B : Sorry for the misunderstanding. Now, about WolframAlpha : The request $x^2 −5x+6=0$ is interpreted as solving the equation for $x$ . Apparently, to make it interpreting as a request to draw $y$ as a function of $x$ , the equation must include $y$ in it. For example, try : http://m.wolframalpha.com/input/?i=plot+%28x%5E2-5x%2B6%29y%3D0&x=0&y=0 End of the copy. Now, if the link doesn't work, type in WolframAlpha : (x^2-5x+6)y=0 and you will see the plot with the two vertical lines. – 2017-02-18