Solution assignment 08 The basics

Solve the following equations:

$(x-3)^2+2(x-3)=0$

$(x-5)^2-16=0$

$x^2+x-1=0$

We can solve the first equation by taking $x-3$ outside brackets. We get:

$(x-3)[(x-3)+2]=0$ , so:

$(x-3)(x-1)=0$ , dus:

$x=3$ of $x=1$

We can write the second equation as:

$x-5=4$ of $x-5=-4$ , so:

$x=9$ of $x=1$

The last equation cannot be solved by factorizion. Then the $abc$ -formula has to be used and always delivers a solution if it exists. We get:

$x=\displaystyle\frac{-1\pm\sqrt{1+4}}{2}$ , and thus:

$x=-\displaystyle\frac{1}{2}+\frac{1}{2}\sqrt{5}$ of $x=-\displaystyle\frac{1}{2}-\frac{1}{2}\sqrt{5}$