Solving quadratic equations by factorization

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.

Product is zero

An equation in the form \[\blue A \cdot \green B =0\]

gives

\[\blue A=0 \lor \green B=0\]

Example

\[ \left(\blue{x-2}\right) \left(\green{x+4}\right)=0 \]

gives

\[\blue{x-2}=0 \lor \green{x+4}=0 \]

Remember that anything multiplied by #0# always equals #0#.

Consider the equation #\blue A \cdot \green B =0#. Here, it is possible to distinguish the following cases.

#\blue A =0# and #\green B=0#

In this case, the equation becomes #\blue0 \cdot \green0 = 0#.

#\blue A =0# and #\green B \ne 0#

In this case, the equation becomes #\blue0 \cdot \green B = 0#.

#\blue A \ne 0# and #\green B =0#

In this case, the equation becomes #\blue A \cdot \green0 =0#.

#\blue A \ne 0# and #\green B \ne 0#

In this case, the equation becomes #\blue A \cdot \green B \ne 0#.

In conclusion, the equation #\blue A \cdot \green B =0# holds if and only if either #\blue A=0# or #\green B=0#.

By applying this theorem, we can quickly solve a quadratic equation that can be factorized.

Solving quadratic equations by factorization

	Procedure	Example
	Solving a quadratic equation for #x# by factorization.	#2x^2+6x+4=6x+6#
Step 1	Reduce the equation until the right-hand side equals #0#.	#2x^2-2=0#
Step 2	Ensure that the coefficient of #x^2# equals #1#.	#x^2-1=0#
Step 3	Factorize the left-hand side of the equation.	#\left(\blue{x+1}\right) \left(\green{x-1}\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{x-1}=0#
step 5	Solve the equations #\blue A=0# and #\green B=0#.	#x=-1 \lor x=1#

Not always applicable

As we saw earlier, not all quadratic expressions can be factorized. Consequently, not every quadratic equation can be solved by following the procedure outlined above. Later, we will introduce a method that works for all quadratic equations. However, this method requires more work, and it is, therefore, advisable to first check if a quadratic equation can be solved by factorization.

Solve #x# using factorization in the equation
\[x^2-14\cdot x+48=0\tiny\]
Give your final answer in the form #x=a\lor x=b#, in which #a# and #b# are numbers.

#x = 8 \lor x = 6#

\[\begin{array}{rcl}
x^2-14\cdot x+48&=&0 \\ &&\phantom{xxx}\blue{\text{original equation}}\\
\left(x-8\right)\cdot \left(x-6\right)&=&0 \\ &&\phantom{xxx}\blue{\text{left-hand side factorized}}\\
x-8=0& \lor& x-6=0 \\ &&\phantom{xxx}\blue{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\
x = 8 &\lor& x = 6 \\ &&\phantom{xxx}\blue{\text{constant terms moved to the right}}\\
\end{array}\]

New example