Why the discriminant determine whether a quadratic has real roots or not?

Question

It's been quiet a mystery for, why is this true:?

If $\Delta>0$ then it have two solutions.

If $\Delta=0$ then it have only one solution.

If $\Delta<0$ then it have no solutions

Answer 1

From the quadratic formula, we have for every quadratic of the form: $$ax^2+bx+c\iff x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ And since the discriminant of a quadratic is $\Delta=b^2-4ac$, we can write $x$ as: $$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}.\tag{$\star$}$$ Now, if $\Delta>0$ then $\sqrt{\Delta}$ exists in $\Bbb R$ and is equal to say $D$, and so there are two distinct solutions, why? Since: $$x_1-x_2=\dfrac{-b+D}{2a}-\dfrac{-b-D}{2a}=\dfrac{D}{a}.$$ If $\Delta=0$ then $\text{($\star$)}$ reduces to $x=\tfrac{-b}{2a}$, which is the only solution. If $\Delta<0$ then the square root of $\Delta$ doesn't exist in $\mathbb R$.

Answer 2

Consider the formula

$$x_{\pm}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$

and moreover $\sqrt{\Delta}$ for $\Delta>0$, $\Delta=0$ and $\Delta<0$.

A more geometric answer is that due to the symmetry of quadratics about the extremum, the extremum is found at the midpoint of the roots, i.e. $\displaystyle \frac{\alpha+\beta}{2}=\frac12 \frac{-b}{a}=-\frac{b}{2a}$ so that

$$a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=-\frac{b^2}{4a}+c$$ is the extremum value of the function.

Now assuming without loss of generality that $a>0$, we have, for example, two real roots if the extremum lies below the $x$-axis aka

$$-\frac{b^2}{4a}+c<0\Rightarrow -b^2+4ac<0\Rightarrow b^2-4ac>0.$$

Answer 3

The mistery can indeed only be resolved when one investigates how one could define discriminant for higher order polynomials. To make a long story short, if a polynomial $$p(x)=x^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0$$ of degree $n$ has $n$ roots $x_1,\ldots, x_n$ one considers the product $$\tag1\Delta=(-1)^{n\choose 2}\prod_{1\le i\le n, 1\le j\le n \atop i\ne j}(x_i-x_j).$$ (Why the factor $(-1)^{n\choose 2}$? This will become clear below!) This expression is symmetric in the $x_i$, that is if we somehow permute the $x_i$, this swaps a few factors and changes a few signs, but everything cancels out so that the product itself is unchanged. Now there is an important theorem saying that any polynomial expression in $n$ variables that is symmetric, can be written as a polynomial expression in the $n$ elementary symmetric polynomials $s_1=x_1+\ldots +x_n$, $s_2=x_1x_2+x_1x_3+\ldots+x_1x_n+x_2x_3+\ldots x_{n-1}x_n$, $\ldots$, $s_n=x_1x_2\cdots x_n$. Finally we observe that for the roots of a polynomial $p$ above we have $s_1=-a_{n-1}$, $s_2=a_{n-2}$, $\ldots$, $s_n=(-1)^na_0$. In other words: $\Delta$ can be expressed as a polynomial in the coefficients of the polynomial $p$. Playing around a bit with the case of quadratic $p$, you may find by yourself that $\Delta=-(x_1-x_2)(x_2-x_1)=x_1^2-2x_1x_2+x_2^2=(x_1+x_2)^2-4x_1x_2=s_1^2-4s_2$.

Now what does $\Delta$ tell about the roots of $p$? First we note that the product $(1)$ is zero if and only if one of the factors is zero, that is: if and only if $p$ has at least one multiple root (in the quadratic case "at least one multiple root" of course means that we have a double root and that's it).

But what does $\Delta>0$ vs. $\Delta<0$ tell us? We can rewrite $(1)$ by combining $(x_i-x_j)$ and $(x_j-x_i)$ into $-(x_i-x_j)^2$; to avoid double counting we consider only the cases with $i<j$: $$ \begin{align}\Delta &=(-1)^{n\choose 2}\prod_{1\le i<j\le n}(-(x_i-x_j)^2)\\&=(-1)^{n\choose 2}\cdot (-1)^{n\choose 2}\prod_{1\le i<j\le n}(x_i-x_j)^2\\&=\left(\prod_{1\le i<j\le n}(x_i-x_j)\right)^2\end{align}$$ That is, $\Delta$ looks like a square. Especially, if all $x_i$ are real, then $\Delta$ is the square of a real number, i.e. nonnegative. On the other hand, each pair of complex conjugate roots gives us a purely imaginary factor, i.e. a sign change for $\Delta$. In the case of quadratic $p$, this is the whole story. For higher degrees, of course the little information contained in the one bit "sign of $\Delta$" may only tell us part of the story.