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Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?

That is, we have one real solution if $$b^2 -4ac = 0,$$ we have two real solutions if $$b^2 -4ac > 0,$$ and we have no real solutions if $$b^2 -4ac < 0.$$

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    $\begingroup$ What sort of an answer to why do you want? It is after all obvious from the formula itself that it behaves as you say (for real coefficients $a, b, c$). $\endgroup$ – PJTraill Oct 12 '17 at 19:17
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Because for $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$ Now, we see that if $b^2-4ac<0$ then $\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}>0$,

which says that the equation $ax^2+bx+c=0$ has no solutions.

For $b^2-4ac=0$ we have one root only: $$x_1=-\frac{b}{2a}$$ and for $\Delta=b^2-4ac>0$ our equation has two distinct roots: $$x_1=\frac{-b+\sqrt{\Delta}}{2a}$$ and $$x_2=\frac{-b-\sqrt{\Delta}}{2a}$$ because in this case we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=$$ $$=a\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)=$$ $$=a\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)=a(x-x_1)(x-x_2).$$

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Think about it geometrically $-$ then compute.

Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ has its apex at $(0,\alpha)$. Looking at the graph as it moves up and down you immediately see how the number of zeros depends on $\alpha$:

  • for $\alpha>0$ we have no zeros.
  • for $\alpha=0$ we have a single zero.
  • for $\alpha<0$ we have two zeros.

Now we can introduct a second paramter $\beta$ to move the parabola left and right: $(x-\beta)^2+\alpha$ has its apex at $(\beta,\alpha)$.

Note: we used the fact that given a function $f(x)$, the graph of the function $f(x-\beta)$ looks exactly like the one of $f$ but shifted to the right by an amount $\beta$.

But of course, shifting a function left and right does not alter the amount of zeros. So it still only depends on $\alpha$. We expand the term a bit:

$$(x-\beta)^2+\alpha=x^2-2\beta x+\beta^2+\alpha.$$

Would your quadratic equation be given in this form, you would immediately see the amount of zeros as describes above. Unfortunately it is mostly given as

$$ x^2+\color{red}px+\color{blue}q=0$$

So instead, you have to look at what parts of the $\alpha$-$\beta$-form above corresponds to these new parameters $p$ and $q$:

$$ x^2\color{red}{-2\beta} x+\color{blue}{\beta^2+\alpha} = 0.$$

So we have $p=-2\beta$ and $q=\beta^2+\alpha$. If we only could extract $\alpha$ from these new parameters, we would immediately see the amount of zeros. But wait! We can!

$$\alpha=q-\beta^2=q-\left(\frac p2\right)^2.$$

This is exactly what you know as (the negative of) the discriminant.


I used the form $x^2+px+q=0$ and you used $ax^2+bx+c=0$. I hope this is not confusing you. Just divide by $a$ (if $a$ is non-zero):

$$x^2+ \color{red}{\frac ba}x+\color{blue}{\frac ca}=0$$

If you set $p=b/a$ and $q=c/a$ and plug this into my discriminant from above you obtain the one you know:

$$\left( \frac {\color{red}p}2 \right )^2-\color{blue}q = \frac{(\color{red}{b/a})^2}4-\color{blue}{\frac ca}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}.$$

Because $4a^2$ is always positive it suffices to look at $b^2-4ac$ as you did in your question.

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    $\begingroup$ Beautifully explained:) $\endgroup$ – Deniz Tuna Yalçın Oct 9 '17 at 13:59
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    $\begingroup$ This is one of the best answers i've seen in this site recently. Have a virtual drink on me. Or an upvote. $\endgroup$ – Mindwin Oct 9 '17 at 17:56
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    $\begingroup$ Love this answer $\endgroup$ – jkd Oct 16 '17 at 11:36
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The discriminant tells you the number of real solutions if $a,b,c$ are real.

That is because $\pm\sqrt{b^2-4ac}$ is real if and only if $b^2-4ac\ge 0.$

If $a,b,c$ are not real then there's more to say than that.

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    $\begingroup$ Thank you for not over-complicating the issue. $\endgroup$ – jpmc26 Oct 9 '17 at 21:03
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Other answers do calculations and it's probably what you asked for, but the answer to "why?" can be deeper, more subtle than that. The discriminant reveals the number of real solutions because:

  • it's useful to know the number of real solutions
  • and at least one formula (function of $a$, $b$ and $c$) that leads to this number does exist.

We discovered (derived) the simplest such formula and said: 'It's so useful it deserves a name, let it be discriminant.' At this point it doesn't matter what the formula is in terms of $a$, $b$ and $c$. I mean, if it were something different than $b^2-4ac$ then this other function would be called 'discriminant' in the first place.

I checked Wikipedia and the first thing I saw was:

In algebra, the discriminant of a polynomial is a polynomial function of its coefficients, which allows deducing some properties of the roots without computing them. […] for a polynomial of an arbitrary degree, the discriminant is zero if and only if it has a multiple root, and, in the case of real coefficients, it is positive if and only if the number of non-real roots is a multiple of 4.

Note few things:

  • If you take a polynomial of degree 2 (so there are two roots, equal or not, either one real or not) with real coefficients, then the above property translates exactly to what you're asking about.
  • Property is the right word here. The above is not a strict definition for sure. $2(b^2-4ac)$ or $(b^2-4ac)^3$ have the same property.
  • Wikipedia gives this rather simple property first, the rather complex general definition later. This suggests it's the property that is important, not the definition itself. The desired property is a reason to have a definition.

In this context, if the question was "why does the value of $b^2-4ac$ reveal the number of real solutions?" then some plain calculations would be the full answer. But the question is "why does the discriminant reveal… ?", so the deeper answer is:

Because we wanted it to reveal this in the first place. We deliberately defined the discriminant to have this property.

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    $\begingroup$ Excellent point, and I've upvoted your answer. People often pick up the wrong end of the stick. (BTW, the spell checker is flagging 'upvoted', in preference to 'up-voted', but I suggest we follow Knuth's suggestion and just drop the hyphen.) $\endgroup$ – Mike Jones Oct 11 '17 at 13:26
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    $\begingroup$ No. The discriminant is a portion of the reduced formula to give any solution to a polynomial quadratic equation. The discriminant is justa thing we took notice of. $\endgroup$ – The Great Duck Oct 11 '17 at 20:56
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    $\begingroup$ Why does the kettle boil? ..... Because electrical energy is converted to heat, and conducted to the water? ..... Or because I want tea? $\endgroup$ – Ben Oct 12 '17 at 12:10
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If the quadratic $ax^2 + bx + c$ has roots $\alpha$ and $\beta$, it factorises as $a(x-\alpha)(x-\beta)$. So we have that in terms of the roots, we can write the coefficients as $b = -a(\alpha + \beta)$, and $c = a \alpha \beta$. Then,

$$ \begin{aligned} \Delta = b^2 - 4ac &= a^2(\alpha^2 + 2 \alpha \beta + \beta^2) - 4a^2\alpha\beta\\ &= a^2(\alpha^2 - 2 \alpha \beta + \beta^2)\\ &= a^2(\alpha - \beta)^2 \end{aligned}$$

So immediately we have that $\Delta = 0$ exactly when $\alpha = \beta$. We can also see that if $\alpha$ and $\beta$ are real, then $\Delta \geq 0$, and if $\Delta < 0$, then $\alpha - \beta$ cannot be a real number.

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    $\begingroup$ You could add that if the roots are not real, they are complex conjugates, so their difference is purely imaginary and its square negative. $\endgroup$ – Carsten S Oct 10 '17 at 15:05
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The roots of $ax^2+bx +c = 0$ are $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

If you graph $ax^2+bx +c = y$, the abcissa of the vertex, the point whose distance between itself and the roots, the points of the parabola that are on the $X$ axis, are equal, and will be $\frac{-b}{2a}$ and it'll be real as long as $b$ and $a$ are real. The distance between the vertex and one of the roots will be $\frac{\sqrt{b^2-4ac}}{2|a|}$. Let us call it $t_a$, the vertex $v_a$ and one of the roots $r_a$, while $point_a$ is the abcissa of $point$; $v_a\pm t_a=r_a$ So if $v_a$ and $t_a$ is real, $r_a$ should be real too. We know that $v_a$ is real as long as the coefficients are real. So $t_a$ should be real too, if we want $r_a$ to be real. To $t_a$ to be real, the condition is, $b^2-4ac$ shouldn't be negative.

The plus-minus sign in $v_a\pm t_a=r_a$ creates two $r_a$ if $t_a$ is nonzero. If $t_a$ is zero, there is only one $r_a$, only one root. If $t_a$ is nonzero and $b^2-4ac$ isn't negative, there are two roots.

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I won't put more formulas here, I'm just trying to explain you the "why" question:

Every quadratic equation can be transformed into a more simplified format:

  • X^2=A

This gives three possibilities:

  1. A>0
  2. A=0
  3. A<0

Each of those is equivalent with a certain amount of real solutions:

  1. Two solutions
  2. One solutions
  3. No

The value A is in line with the discriminant.

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I like to think of it this way. The axis of symmetry for the parabola is

-b/2a

That is not-coincidentally part of the quadratic formula. The remining poriton of the formula indicates the x values of the roots on either side of that axis. This also not-coincidentally contains the determinant

 ±√(b^2-4ac)/2a  --> √(b^2-4ac)

So, you have a formula which describes how far on either side of the axis one will find the roots of the parabola. How does this formula work? The main feature is a square root which is defined on the domain

D:{x|x∈ℝ && x≥0)

And within that domain we have the feature that for x=0, there is one value √0=0 BUT for any number x>0 there are two values e.g. √4 = +2 and -2.

So, now that we have covered all the background information the answer to your problem is pretty simple. There are three kinds of answers from the determinant because there are three portions of ℝ in the domain which produce different results.

 x<0  --> No real solution
 x=0  --> Exactly one solution
 x>0  --> Two solutions, positive and negatives integers of the same absolute value

In the determinant we can easily determine which of the three cases we have. In the full quadratic formula we end up with

No real root on either side of the axis of symmetry
One real root on the apex of the parabola and on the axis of symmetry
Two real roots equidistance on either side of the axis of symmetry

Graphic showing the three cases

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Say your quadratic equation is $$ ax^2 + bx +c = 0 $$ and suppose that $a>0$ , so that we can imagine $p(x) = ax^2 + bx +c$ to be a parabola which is open at the top. Let's call $m$ the minimum value of $p(x)$ . Graphically it is just the lowest point of the parabola.

If you now draw a few examples for parabolas, you will notice that, depending on the sign of $m$, the parabola will cut, touch or never touch the horizontal line ($x$-axis). Depending on the situation, you can see how many roots there are: \begin{align*} &\text{if} ~~m>0~~ \text{we don't touch the line $\longrightarrow$ no roots} \\ &\text{if} ~~m = 0~~ \text{we touch the line in one point $\longrightarrow$ one root} \\ &\text{if} ~~m < 0~~ \text{we cross the line $\longrightarrow$ two roots} \end{align*}
To find the $x$ coordinate of our minimum $m$, we can use the derivative, since it has to be $0$ there. $$ 0 = p'(x) = 2ax + b ~~~\Longleftrightarrow ~~~ x = -\frac{b}{2a} $$ so we get $$ m = p \left( -\frac{b}{2a} \right) = a \frac{b^2}{4a^2} - b\frac{b}{2a} +c = \frac{-b^2 + 4ac}{4a} $$ Since we are only interested in the sign of $m$, and since $a>0$ , the sign of $m$ will be the same as the sign of $-b^2 +4ac$ . If we combine this with earlier observation, we have \begin{align*} &\text{if} ~~~4ac - b^2 >0~~~ \text{we don't touch the line $\longrightarrow$ no roots} \\ &\text{if} ~~~4ac - b^2 = 0~~~ \text{we touch the line in one point $\longrightarrow$ one root} \\ &\text{if} ~~~4ac - b^2 < 0~~~ \text{we cross the line $\longrightarrow$ two roots} \end{align*}

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