The general form of a quadratic equation is as follows:

$ax^2+bx+c = 0$

Examples of quadratic equations:

- $x^2+2x+1=0$

Using factorisation, $x^2+2x+1=0$ can be written as:

$$(x+1)(x+1)=0$$

Using the rule that $a*0 = 0$ for all $a \in \mathbb{R}$ meaning that when two numbers or expressions multiply each other, then one of them must be equal to zero.

In this case the first $x+1=0$ or the second $x+1=0$

therefore the solution is $x={-1}$ twice.

2. $5x^2+6x+1=0$

Using factorisation, $5x^2+6x+1=0$ can be written as:

$$(x+5)(x+1)=0$$

Using the rule that $a*0 = 0$ for all $a \in \mathbb{R}$ meaning that when two numbers or expressions multiply each other, then one of them must be equal to zero.

In this case the first $x+5=0$ or the second $x+1=0$

therefoere the solution is $x={-5}$ or $x={-1}$

3. $(x-3)(x+4)=0$

$(x-3)=0$ or $(x+4)=0$

$x=3$ or $x={-4}$

$ x(x+7)=0$

$x=0$ or $x+7=0$

$x=0$ or $x={-7}$

$x^2=9$$

Using difference of two squares, $x^2=9 = (x-3)(x+3)=0$

$$x-3=0$$ or $$x+3=0$$

$$x=3$$ or $$x={-3}$$

At times when solving quadratic equations it will be impossible to factorise thus it is necessary to use the $\textbf{Quadratic Formula}$.

Using the general form of a quadratic equation: $ax^2+bx+c = 0$ arrange your equation in the same way with the right hand side having 0, then identify the values of $a,b$ and $c$ e.g. in $5x^2+6x+1=0$, $a=5, b=6$ and $c=1$ and substitute them in the quadratic formula.

The $\textbf{Quadratic Formula}$ is as follows:

$x = \frac{{-b}\pm{\sqrt{b^2 – 4ac}}}{2a}$

**Example:**

$$5x^2+6x+1=0$$, $$a=5, b=6$$ and $$c=1$$

$$x = \frac{{-6}\pm{\sqrt{{-6}^2 – 4(5)(1)}}}{2(5)}$$

Solution: $x= \frac{-1}{5}$ or $x= {-1}$

**Exercise**

Solve each of the following quadratic equations both by factorisation and quadratic formula.

$$(x+5)(x-3)=0$$

$$(x-2)(x-1)$$

$$(x+1)^2=0$$

$$(x+5)(3x-1)=0$$

$$(2x+1)((3x+2)=0$$

$$8x=12x^2$$

$$3x^2-15x=108$$

$$2(x-2)(x+4)=32$$

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