### Basic functions: Polynomial functions

### The quadratic formula

Let \(a\), \(b\), and \(c\) be real numbers with \(a\neq 0\).

The **discriminant** of the quadratic equation \(ax^2+bx+c = 0\) is defined as the number \(b^2-4\cdot a \cdot c\).

The reason for introducing the discriminant (hereafter denoted by the letter \(D\)) is that we can now formulate a rule for how many real solutions the quadratic equation has and, if solutions exist, which exact values they have.

The formula below, which directly gives the solutions and their number, is called the **quadratic formula.**

The quadratic formula The quadratic equation \(ax^2+bx+c = 0\) with unknown \(x\) and discriminant \(D=b^2-4ac\) has:

- two real solutions if \(D\gt 0\), namely \(x=\dfrac{-b - \sqrt{D}}{2a}\) and \(x=\dfrac{-b+ \sqrt{D}}{2a}\).
- exactly one real solution if \(D=0\), namely \(x=-\dfrac{b}{2a}\).
- no real solutions if \(D\lt 0\).

The two solutions in the first case are often written together, by making use of the \(\pm\) notation; so \[x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\]

*gives as a solution \[x=\dfrac{-b\pm\sqrt{D}}{2a}\] with \[D=b^2-4ac\tiny.\] In this problem with the equation \[2\, x^2-8\, x+4=0\] we have: \[a=2,\quad b=-8,\quad c=4,\quad D=(-8)^2 -4\times 2\times 4= 32\tiny.\] Because \(D\gt 0\), the number of solutions of the equation \(2\, x^2-8\, x+4=0\) is equal to \(2\).*

The answer is according to the quadratic formula (possibly, after simplification): \[ x=2-\sqrt{2}\quad\vee\quad x=2+\sqrt{2}\tiny.\]