Quadratic equations: Solving quadratic equations
Solving quadratic equations by factorization
Earlier, we learned how to do single brackets and double brackets factorization. We will now apply these skills to solve quadratic equations.
An equation in the form \[\blue A \cdot \green B =0\]
gives
\[\blue A=0 \lor \green B=0\]
Example
\[ \left(\blue{x2}\right) \left(\green{x+4}\right)=0 \]
gives
\[\blue{x2}=0 \lor \green{x+4}=0 \]
By applying this theorem, we can quickly solve a quadratic equation that can be factorized.
Procedure 
Example 

Solving a quadratic equation for #x# by factorization. 
#2x^2+6x+4=6x+6# 

Step 1 
Reduce the equation until the righthand side equals #0#. 
#2x^22=0# 
Step 2 
Ensure that the coefficient of #x^2# equals #1#. 
#x^21=0# 
Step 3 
Factorize the lefthand side of the equation. 
#\left(\blue{x+1}\right) \left(\green{x1}\right)=0# 
step 4 
Applying the rule #\blue A \cdot \green B =0# gives #\blue A=0 \lor \green B=0#. 
#\blue{x+1}=0 \lor \green{x1}=0# 
step 5 
Solve the equations #\blue A=0# and #\green B=0#. 
#x=1 \lor x=1# 
#\begin{array}{rcl}
x^28\cdot x+7&=&0 \\ &&\phantom{xxx}\blue{\text{original equation}}\\
\left(x7\right)\cdot \left(x1\right)&=&0 \\ &&\phantom{xxx}\blue{\text{lefthand side factorized}}\\
x7=0& \lor& x1=0 \\ &&\phantom{xxx}\blue{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\
x = 7 &\lor& x = 1 \\ &&\phantom{xxx}\blue{\text{constant terms moved to the right}}\\
\end{array}
#
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