Quadratic equations: Solving quadratic equations
Quadratic equations
We are interested in the solutions of 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
#none#
#\begin{array}{rclcl}x^2+26&=&21\\&&\phantom{xxx}\blue{\text{the given equation}}\\ x^2&=&-5\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}}\\ none&&\\&&\phantom{xxx}\blue{\text{according to the theory}}\end {array}#
#\begin{array}{rclcl}x^2+26&=&21\\&&\phantom{xxx}\blue{\text{the given equation}}\\ x^2&=&-5\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}}\\ none&&\\&&\phantom{xxx}\blue{\text{according to the theory}}\end {array}#
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