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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    $\begingroup$ @lamar Thanks for the edit. I really should have included the graph earlier, stupid me. $\endgroup$ – A---B Feb 18 '17 at 5:22
  • $\begingroup$ 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). $\endgroup$ – jacobq Oct 25 '18 at 18:29

$x^2-5x+6=0\quad$ 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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    $\begingroup$ $y^2=x^2$ 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 $\left( y(x)=x \text{ and } y(x)=-x \right)$. $\endgroup$ – JJacquelin Feb 17 '17 at 7:44
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    $\begingroup$ @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: $\{ (x,y)\mid x^2-5x+6=0 \}$ $\endgroup$ – P Vanchinathan Feb 17 '17 at 7:51
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    $\begingroup$ @ 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" . $\endgroup$ – JJacquelin Feb 17 '17 at 8:07
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    $\begingroup$ 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. $\endgroup$ – Mike Feb 17 '17 at 8:27
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    $\begingroup$ @ 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. $\endgroup$ – JJacquelin Feb 17 '17 at 9:23

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.

  • $\begingroup$ But when I enter wolframalpha.com/input/?i=x%5E2+%2B+4xy+%2B+y%5E2+%3D+0 I do get a pair of striaght lines. $\endgroup$ – A---B Feb 17 '17 at 8:04
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    $\begingroup$ Because you provided independent variable $x$ and $y$ and not the same $x$ twice. $\endgroup$ – mvw Feb 17 '17 at 8:08
  • $\begingroup$ @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. $\endgroup$ – skyking Feb 17 '17 at 8:12
  • $\begingroup$ @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). $\endgroup$ – skyking Feb 17 '17 at 8:15
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    $\begingroup$ @A---B Yes, but if you study the 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). $\endgroup$ – skyking Feb 17 '17 at 9:20

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.

  • $\begingroup$ 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. $\endgroup$ – JJacquelin Feb 17 '17 at 10:07

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".

  • $\begingroup$ Ok, but consider "x^2 -4xy + y^2 = 0$ which actually when plotted on WA gives a pair of lines, What do you think ? $\endgroup$ – A---B Feb 17 '17 at 18:00
  • $\begingroup$ $x^2 -4xy +y^2 = 0$ 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. $\endgroup$ – Paul Sinclair Feb 17 '17 at 22:18
  • $\begingroup$ 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. $\endgroup$ – A---B Feb 18 '17 at 4:38
  • $\begingroup$ 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. $\endgroup$ – Paul Sinclair Feb 18 '17 at 17:08

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.

  • $\begingroup$ @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. $\endgroup$ – Mike Feb 17 '17 at 7:24
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    $\begingroup$ 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 $3$. $\endgroup$ – Mike Feb 17 '17 at 7:34
  • $\begingroup$ I made the interpretations explicit. $\endgroup$ – mvw Feb 17 '17 at 8:11
  • $\begingroup$ 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? $\endgroup$ – Mike Feb 17 '17 at 8:42
  • $\begingroup$ @mvw But can you tell why WA differentiate between $x^2-5x+6 =0$ and $x^2 -4xy +y^2 = 0$ ? $\endgroup$ – A---B Feb 17 '17 at 9:07

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 $y=x^2 - 5x + 6$.

  • $\begingroup$ 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. $\endgroup$ – Jeppe Stig Nielsen Feb 17 '17 at 13:02

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.


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

  • $\begingroup$ The example of Plot y(x^5-5x+6)=0 in WolftanAlpha has been cited in comments a long time ago. $\endgroup$ – JJacquelin Feb 18 '17 at 17:00
  • $\begingroup$ @JJacquelin I have read every single comment but I can't find it. Can you please show me where you have seen it ? $\endgroup$ – A---B Feb 18 '17 at 17:56
  • $\begingroup$ 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 : m.wolframalpha.com/input/… 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. $\endgroup$ – JJacquelin Feb 18 '17 at 18:09

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