# Mathematics for GIS Professionals

The general form of a quadratic equation is as follows:

$ax^2+bx+c = 0$

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

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=0x=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=1x = \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)=08x=12x^23x^2-15x=1082(x-2)(x+4)=32$\$

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