#### Chapter 2 Equations in one Variable

Section 2.1 Simple Equations

##### Info 2.1.20

A quadratic equation is an equation of the form $a{x}^{2}+bx+c=0$ with $a\ne 0$, or, in reduced form, ${x}^{2}+px+q=0$. This form is obtained by dividing the equation by $a$.
For a quadratic equation in one variable (here the variable $x$) one of the following three statements holds:
• The quadratic equation has no solution: $L=\left\{\right\}$.
• The quadratic equation has a single solution $L=\left\{{x}_{1}\right\}$.
• The quadratic equation has two different solutions $L=\left\{{x}_{1};{x}_{2}\right\}$.

The solutions are obtained by applying quadratic solution formulas.
##### Info 2.1.21

The $pq$ formula for solving the equation ${x}^{2}+px+q=0$ reads

${x}_{1,2}\mathrm{ }=\mathrm{ }-\frac{p}{2}±\sqrt{\frac{1}{4}{p}^{2}-q} .$

Here, the equation has
• no (real) solution if $\frac{1}{4}{p}^{2}-q<0$ (taking the square root is not allowed),
• a single solution ${x}_{1}=-\frac{p}{2}$ if $\frac{1}{4}{p}^{2}=q$ and the square root is zero,
• two different solutions if the square root is a positive number.

The expression $D:=\frac{1}{4}{p}^{2}-q$ underneath the square root considered above is called the discriminant.
The solution of a quadratic equation is often described by an alternative formula:
##### Info 2.1.22

For the equation $a{x}^{2}+bx+c=0$ with $a\ne 0$ the $abc$ formula reads

${x}_{1,2}\mathrm{ }=\mathrm{ }\frac{-b±\sqrt{{b}^{2}-4ac}}{2a} .$

Here, the equation has
• no (real) solution if ${b}^{2}-4ac<0$ (the square root of a negative number is undefined within the range of real numbers),
• a single solution ${x}_{1}=-\frac{b}{2a}$ if ${b}^{2}=4ac$ and the square root is zero,
• two different solutions if the square root is a positive number.
Again, the expression $D:={b}^{2}-4ac$ underneath the square root considered above is called the discriminant.
Both formulas result in the same solutions. The $\mathrm{pq}$ formula is easier to learn, but is only applicable if $a$, the coefficient of the quadratic term, is $1$. Otherwise we must divide both sides of the equation by $a$.

In terms of the $\mathrm{pq}$ formula, the three different cases correspond to three possibilies for the number of intersection points that the graph of a shifted standard parabola $f\left(x\right)={x}^{2}+px+q$ may have with the $x$ axis.

Three cases: no intersection point, one intersection point, two intersection points with the $x$ axis.
##### Example 2.1.23
The quadratic equation ${x}^{2}-x+1=0$ has no solution since the discriminant $\frac{1}{4}{p}^{2}-q=-\frac{3}{4}$ within the $pq$ formula is negative. In contrast, the equation ${x}^{2}-x-1=0$ has two solutions

$\begin{array}{ccc}\multicolumn{1}{c}{{x}_{1}}& =\hfill & \frac{1}{2}+\sqrt{\frac{1}{4}+1}\mathrm{ }=\mathrm{ }\frac{1}{2}\left(1+\sqrt{5}\right) , \hfill \\ \multicolumn{1}{c}{{x}_{2}}& =\hfill & \frac{1}{2}-\sqrt{\frac{1}{4}+1}\mathrm{ }=\mathrm{ }\frac{1}{2}\left(1-\sqrt{5}\right) .\hfill \end{array}$

##### Info 2.1.24

The function expression of a parabola has vertex form if the function has the form $f\left(x\right)=a·\left(x-s{\right)}^{2}-d$ with $a\ne 0$. In this case, $\left(s;-d\right)$ is the vertex of the parabola. The corresponding quadratic equation for $f\left(x\right)=0$ then reads $a·\left(x-s{\right)}^{2}=d$.
Dividing this equation by $a$ one obtains the equivalent quadratic equation $\left(x-s{\right)}^{2}=\frac{d}{a}$. Since the left-hand side is a square of a real number, only solutions exist if and only if the right-hand side is non-negative as well, i.e. $\frac{d}{a}\ge 0$. By taking the square root, taking the two possible signs into account, one obtains $x-s=±\sqrt{\frac{d}{a}}$.
So, for $\frac{d}{a}>0$ two solutions of the equation exist:

${x}_{1}\mathrm{ }=\mathrm{ }s-\sqrt{\frac{d}{a}} , {x}_{2}\mathrm{ }=\mathrm{ }s+\sqrt{\frac{d}{a}} ;$

they are symmetric to the $x$ coordinate $s$ of the vertex. For $d=0$, only one solution exists.
The sign of $a$ determines whether the function expression describes a parabola opening upwards or downwards.

The quadratic equation has only one single solution $s$ if it can be transformed into the form $\left(x-s{\right)}^{2}=0$.
##### Info 2.1.25

Any quadratic equation can be transformed (after collecting terms on the left-hand side and normalisation, if necessary) to vertex form by completing the square. For this, a constant is added to both sides of the equation such that on the left-hand side we have a term of the form ${x}^{2}±2sx+{s}^{2}$ to which the first or second binomial formula can be applied.

##### Example 2.1.26
Adding the constant $2$ transforms the equation ${x}^{2}-4x+2=0$ into the form ${x}^{2}-4x+4=2$ or into the form $\left(x-2{\right)}^{2}=2$, respectively. From this, the two solutions ${x}_{1}=2-\sqrt{2}$ and $2+\sqrt{2}$ can be seen immediately. In contrast, the quadratic equation ${x}^{2}+x=-2$ has no solution since completing the square results in ${x}^{2}+x+\frac{1}{4}=-\frac{7}{4}$ or $\left(x+\frac{1}{2}{\right)}^{2}=-\frac{7}{4}$, respectively, where the right-hand side is negative for $a=1$.

##### Exercise 2.1.27
Find the solutions of the following quadratic equations by completing the square after collecting terms on the left-hand side and normalisation (i.e. selecting $a=1$):
1. ${x}^{2}=8x-1$ has the vertex form
$=$ .
The solution set is $L$$=$ .
2. ${x}^{2}=2x+2+2{x}^{2}$ has the vertex form
$=$ .
The solution set is $L$$=$ .
3. ${x}^{2}-6x+18=-{x}^{2}+6x$ has the vertex form
$=$ .
The solution set is $L$$=$ .
Enter sets in the form $\left\{$ a;b;c;$\dots \right\}$. Enter the empty set as $\left\{\right\}$.