Below everything is explained but I would advise working through the task above and look at the videos when stuck.

solution videos for the above task.

It is possible to write every quadratic equation in the form

$${ax^2 + bx + c = 0}$$, where a is a real number not equal to 0 and b and c are real numbers.

The constants a, b and c are called coefficients (Koeffizienten). In German textbooks, when ${a = 1}$, the quadratic equation is in the ‘so-called’ Normalform - $${ x^2 + px + q = 0}$$

## no linear term in the quadratic equations - Die reinquadratische Gleichung

$${ax^2 + c = 0}$$

If ${b = 0}$, the linear term ${bx}$ in the quadratic equation is missing. It then has the form

{\begin{align} ax^2 + c & = 0 \newline ax^2 & = - c \newline x^2 & = - \frac{c}{a} \newline x^2 & = q \end{align}}

#### Example 1

{\begin{align}5x^2 - 80 & =0\newline 5 \cdot x^2 - 80 & = 0 \newline x^2 - 16 & = 0\newline x^2 & = 16\newline x & = \pm \sqrt{16}\end{align}} $${x_1 = 4 ~ \lor x_2 = -4}$$

${ \lor}$ means OR.

The number of solutions is determined as follows.

• one solution (eine Lösung) for ${q = 0}$
• two solutions (zwei Lösungen) for ${q > 0}$
• no solution (keine Lösung) for ${q < 0}$

## The constant term is zero - Die Konstante ist Null

When the constant term is zero, i.e., ${c = 0}$, the quadratic equation has the following form $${ax^2 + bx = 0}$$ where a is a real number not equal to 0 and b is any real number. We solve these equations by factorising (Questions 1 and 3 only). {\begin{align}ax^2 + bx & =0 \newline x \cdot (ax + b) & = 0 \end{align}} applying the Null Factor Law which states that if the product is zero then at least one of the factors is zero.

$${x= 0 \lor ax + b = 0}$$ $${ ax = -b}$$ $${ x = -\frac{b}{a}}$$ $${x_1 = 0 ~ \lor x_2 = -\frac{b}{a}}$$

All quadratic equations of the form ${ax^2+bx=0}$ have a common solution ${x=0}$.

#### Example 2

{\begin{align}5x^2 + 2x & = 0 \newline x \cdot (5x + 2) & = 0 \end{align}} It follows that {\begin{align}x= 0 ~~ \lor 5x + 2 & = 0 \newline \Leftrightarrow 5x & = -2 \newline \Leftrightarrow x & = -\frac{2}{5}\end{align}} $${x_1 = 0~ \lor x_2 = - \frac{2}{5} }$$

For more practice go to following links

## factorising (theorem of Viète) - Satz von Vieta1

When the quadratic equation is in factored form $${(x-x_1)\cdot(x-x_2) = 0}$$

the solutions can be found using the null factor law. Set each factor equal to zero.

{\begin{align} x-x_1 & = 0 &&\lor &&& x-x_2 = 0 \newline x & = x_1 &&\lor &&& x = x_2 \end{align}} $${\mathbb{L} = \{ x_1; x_2 \}}$$

The theorem of Viète indicates if the quadratic equation can be factorised. If a quadratic equation ${x^2 + bx + c = 0}$ has two solutions ${x_1}$ and ${ x_2}$, then following holds - $${x^2 + bx + c =(x - x_1) \cdot (x - x_2)}$$ where

$${x_1 + x_2 = -b}$$ $${x_1 \cdot x_2 = c}$$

#### Example 3

Factorise $${ x^2 - 6x + 8 = 0}$$ First, we have b = -6 and c = 8. According to Vieta following statement must hold. $${c = x_1 \cdot x_2 = 8 }$$ So, find all the combinations of factors of 8. $${ 8 = 8 \cdot 1 = (-8)\cdot(-1)=4 \cdot 2 = (-4)\cdot(-2)}$$ Which of these factors also holds for Vieta’s second statement? $${x_1 + x_2 = -b = -(-6)= 6}$$ The only two factors giving -b = 6 are $${x_1 = 4 }$$ and $${x_2 = 2}$$ Hence, $${ x^2 - 6x + 8 = (x - 4)\cdot (x-2)}$$ And, we can rewrite the equation in factored form. $${(x - 4)\cdot (x-2)=0}$$ $${\Leftrightarrow x-4 = 0 \lor x -2 = 0}$$ $${\Leftrightarrow x_1 = 4 \lor x_2 = 2}$$ So, the solution set is- $${\mathbb{L} = \{ 2; 4 \}}$$

## Completing the square - quadratische Ergänzung

Completing the square means to rearrange part of the equation into a perfect square. $${x^2 + 2ux + u^2 = (x + u)^2}$$

#### Example 4

{\begin{align} \frac{1}{2}x^2 +5x & = 28 && \text{subtract 28} \newline \frac{1}{2}x^2 +5x - 28 & = 0 && \text{multiply by 2}\newline x^2 + 10x - 56 & = 0 && \text{eliminate constant term on left hand side} \newline x^2 + 10x & = 56 && \text{complete the square}\end{align}} $${2ux = 10x => u = 5 => u^2 = 25}$$ {\begin{align} x^2 + 10x + 25 & = 56 + 25 && \text{binomial formula} \newline (x + 5)^2 & = 81 && \text{square root}\newline x + 5 & = \pm 9 \end{align}} $${ x + 5 = + 9 \lor x + 5 = - 9 }$$ $${ x = -5 + 9 \lor x = -5 - 9 }$$ $${ x = 4 \lor x = -14 }$$ $${\mathbb{L} = \{ 4; -14 \} }$$

To solve for roots in the general form see this interactive page.

Following link on completing the square has many exercises. The approach is slightly different to the one explained above as all terms remain on the left hand side. The method on this page saves some steps.

Completing the square for the expression $${ax^2+bx+c=0}$$ leads to what is known as the quadratic formula.

If ${b^2-4ac > 0}$, the quadratic formula to solve ${ax^2+bx+c=0}$ is

$${x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

#### examples on video

Here are two videos showing you how to use the quadratic formula.

See no 14 on the solution page The notation ${x_{1,2} }$ means there are two solutions ${x_1 }$ and ${x_2}$.

But, if $${b^2-4ac = 0}$$, there is one solution:$${x = \frac{-b}{2a}}$$

And lastly, if ${b^2-4ac < 0}$, the square root of a negative number is not defined for real numbers. There is no real solution.

Note, the expression ${b^2-4ac}$ is also called the discriminant ${\Delta}$.

(later on, you may come across complex numbers which provide solutions for negative roots).

## The pq - formula - a special case

(used in some parts of Germany, especially in Berlin)

If the equation is given in the following form: $${ x^2 + px + q = 0}$$

Then the quadratic equation is in ‘normalised’ form, i.e., the coefficient of the quadratic term is 1, i.e., a = 1 , then the quadratic formula is simplified to what is known as the p-q-formula.

$${x_{1,2} = \frac{-p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}}$$

See no 13b on the following video

#### examples on video

Here are two videos showing you how to use the pq-formula.

#### “completing the square” to derive the p-q-formula - Herleitung der pq-Formel

{\begin{align} x^2 + px + q & = 0 \newline x^2 + px & = -q \newline x^2 + px + \left(\frac{p}{2}\right)^2 & = -q + \left(\frac{p}{2}\right)^2 \newline \left(x + \left(\frac{p}{2}\right) \right)^2 & = \left(\frac{p}{2}\right)^2 -q \newline x + \left(\frac{p}{2}\right) & = \pm \sqrt{ \left(\frac{p}{2}\right)^2 -q} \newline x & = - \left(\frac{p}{2}\right) \pm \sqrt{ \left(\frac{p}{2}\right)^2 -q} \end{align}}

You can convert all quadratic equations into the p-q-form by dividing all terms by a (the coefficient of the quadratic term)

### Solving quadratics where it is necessary to investigate different cases

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