### Quadratic equations: Solving quadratic equations

### Quadratic equations

We are interested in the solutions of a quadratic equations, these are equations of the form #ax^2+bx+c=0#. There are several ways to solve such an equation. We will start out by solving the simplest quadratic equation #x^2=\blue c#.

If #\blue c>0#, the equation has *two solutions*:

\[\begin{array}{rcl}

x^2 &=&\blue c \\

&\text{gives}&\\

x = \sqrt{\blue c} &\lor& x = -\sqrt{\blue c}

\end{array}\]

The symbol #\lor# means "or".

geogebra picture

If #\blue c=0#, the equation has *one solution*:

\[\begin{array}{rcl}

x^2 &=& \blue c \\

&\text{gives}&\\

x &=&0

\end{array}\]

geogebra picture

If #\blue c<0#, the equation has *no solutions*:

\[\begin{array}{rcl}

x^2 &=&\blue c

\end{array}\]

The graph on the right demonstrates that the graphs #y=\blue c# and #y=x^2# never intersect as long as #\blue c<0#.

geogebra picture

#z=0#

#\begin{array}{rclcl}z^2+40&=&40\\&&\phantom{xxx}\blue{\text{the given equation}}\\ z^2&=&0\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}}\\ z&=&0\\&&\phantom{xxx}\blue{\text{according to the theory}}\end {array}#

#\begin{array}{rclcl}z^2+40&=&40\\&&\phantom{xxx}\blue{\text{the given equation}}\\ z^2&=&0\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}}\\ z&=&0\\&&\phantom{xxx}\blue{\text{according to the theory}}\end {array}#

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