Hi. This page will be about the quadratic formula which is commonly taught in high school mathematics (in Canada).
What Is The Quadratic Formula?
Deriving The Quadratic Formula
The Discriminant And Three Cases
Using The Quadratic Formula Through Examples
The quadratic formula is a useful formula for solving x-intercepts of quadratic equations in the form of
\[ y = ax^2 + bx + c\]
The quadratic formula (with \(a \neq 0\)) is:
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
It is preferable to use the quadratic formula when factoring techniques do not work.
For many (highschool) students, it is not expected to know how to come up with the quadratic formula. This proof is more of a reference to help you see where this formula comes from.
We start with the quadratic form of \(y = ax^2 + bx + c\). Factoring the first two terms by \(a\) gives us:
\[ y = a (x^2 + \dfrac{b}{a}x) + c \]
The coefficient that is with \(x\) is \(\dfrac{b}{a}\). This \(\dfrac{b}{a}\) is divided by 2 gives us \(\dfrac{b}{2a}\). The square of \(\dfrac{b}{2a}\) yields \((\dfrac{b}{2a})^2\) or \(\dfrac{b^2}{4a^2}\). Adding and subtracting by \(\dfrac{b^2}{4a^2}\) gives us:
\[y = a (x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2} - \dfrac{b^2}{4a^2} ) + c\]
We then factor the first three terms. That is, we factor \(x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2}\) into \((x + \dfrac{b}{2a})^2\).
\[y = a ((x + \dfrac{b}{2a})^2 - \dfrac{b^2}{4a^2}) + c \qquad \text{(Factoring the first three terms.)} \]
The distributive law is applied such that \(a\) is expanded to the first two terms.
\[y = a(x + \dfrac{b}{2a})^2 - \dfrac{ab^2}{4a^2} + c \qquad \text{(Expand $a$ to the first two terms.)} \]
Next, we simplify the \(\dfrac{ab^2}{4a^2}\) term.
\[y = a(x + \dfrac{b}{2a})^2 - \dfrac{b^2}{4a} + c \]
Because we want to find x-intercepts, we set \(y = 0\) and solve for \(x\).
\[ 0 = a(x + \dfrac{b}{2a})^2 - \dfrac{b^2}{4a} + c \]
\[ a(x + \dfrac{b}{2a})^2 = \dfrac{b^2}{4a} - c \qquad \text{(Rearranging)}\]
\[(x + \dfrac{b}{2a}) = \pm \sqrt{\dfrac{b^2}{4a^2} - \dfrac{c}{a}} \qquad \text{(Divide by $a$ and take the square root on both sides.)}\]
\[x = -\dfrac{b}{2a} \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} \qquad \text{(Rearrange and find common denominator.)}\]
\[x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \qquad \text{(Square root of top and bottom)}\]
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The final line is the quadratic formula or the \(x\) value such that it makes \(y = 0\) in \(y = ax^2 + bx + c\).
Notice how in the quadratic formula there is a square root part after the plus and minus sign (\(\pm\)). The part inside the square root (\(b^2 - 4ac\)) is called the discriminant.
An important property of square roots is that square roots take on numbers which are at least 0 (non-negative). A negative number inside the square root is undefined (in the real numbers).
We look at three cases for the discriminant and what each case means.
If \((b^2 - 4ac) > 0\) then there would be 2 distinct solutions for \(x\) (or x-intercepts) in the equation $ 0 = ax^2 + bx + c$.
If \((b^2 - 4ac) = 0\) then there would be one value for \(x\) in the equation \(0 = ax^2 + bx + c\).
If \((b^2 - 4ac) < 0\), we would have a negative value inside the square root. The square root of a negative value is undefined. There would be no real-numbered values for \(x\) in the equation \(0 = ax^2 + bx + c\).
The quadratic formula can be applied to any quadratic equation in the form \(y = ax^2 + bx + c\). It does not really matter whether the quadratic form can be factored or not.
Example One
Given the quadratic equation \(y = x^2 + x - 1\), what are the x-intercepts?
From \(y = x^2 + x - 1\), we have \(a = 1\), \(b = 1\) and \(c = -1\). Using these values, the quadratic formula is as follows:
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}\]
\[x = \dfrac{-1 \pm \sqrt{1 + 4}}{2}\]
\[x = \dfrac{-1 \pm \sqrt{5}}{2}\]
The x-intercepts (when \(y = 0\)) for \(y = x^2 + x - 1\) are \(x = \dfrac{-1 + \sqrt{5}}{2}\) (or \(\approx\) 0.618034) and \(x = \dfrac{-1 - \sqrt{5}}{2}\) (or \(\approx\) -1.618034). You can choose to use exact values or using the decimal of the answers (with a calculator).
Example Two
Apply the quadratic formula to find x-intercepts for the equation \(y = 2x^2 -3x - 1\).
Here, we have \(a = 2\), \(b = -3\) and \(c = -1\).
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)}\]
\[x = \dfrac{3 \pm \sqrt{9 + 8}}{4}\]
\[x = \dfrac{3 \pm \sqrt{17}}{4}\]
The x-intercepts are \(x = \dfrac{3 + \sqrt{17}}{4}\) (or \(\approx\) 1.780776) and \(x = \dfrac{3 - \sqrt{17}}{4}\) (or \(\approx\) -0.2807764).
Example Three
Show that the equation \(y = x^2 -2x + 5\) has no x-intercepts.
You can use the quadratic formula right away here. However, it is a bit easier and faster to check the discriminant \((b^2 - 4ac)\).
\[b^2 - 4ac = (-2)^2 -4(1)(5)\]
\[b^2 - 4ac = 4 - 20 \]
\[b^2 - 4ac = -16 \]
\[b^2 - 4ac < 0 \]
Since the discriminant \((b^2 - 4ac)\) is negative, we have a negative inside the square root of the quadratic formula. The equation \(y = x^2 -2x + 5\) has no x-intercepts.