Solving equations with factorization

Functions: Quadratic functions

Solving equations with factorization

If we can factor the left hand side of the equation #ax^2+bx+c=0#, then the solution is easily found:

Solving a quadratic equation by factorization Consider the quadratic equation with unknown #x#: \[ax^2+bx+c=0\tiny,\] where #a#, #b#, and #c# are real numbers with #a\ne0#. If the left hand side can be factored in linear factors: \[ ax^2+bx+c=(x-p)\cdot(x-q)\tiny,\] where #p# and #q# are real numbers, then the solution of the equation is #x=p\lor x=q#.

Here we used the following rule \[A\cdot B=0\] \[A=0\lor B=0\tiny,\] in which #A = x-p# and #B = x-q#:

\[\begin{array}{rcl}ax^2+bx+c&=&0\\ &&\phantom{xx}\color{blue}{\text{original equation}}\\ (x-p)\cdot(x-q)&=&0\\&&\phantom{xx}\color{blue}{\text{left hand side replaced by factorization}}\\ x-p=0&\lor& x-q=0\\&&\phantom{xx}\color{blue}{\text{above mentioned rule}}\\ x=p&\lor& x=q \\&&\phantom{xx}\color{blue}{\text{constants to right hand side}}\\ \end{array}\]

We apply this observation to the following example.

Solve the equation for #x# below by using factorization.
\[x^2-4 x-15=-6 x\tiny.\]
Give your answer in the form #x=a\lor x=b#.

#x=-5\lor x=3#

This can be seen by means of the following deduction.
\[\begin{array}{rcl}
x^2-4 x-15&=&-6 x \\
&&\phantom{xxx}\color{blue}{\text{the original equation}}\\
x^2+2 x-15&=&0 \\
&&\phantom{xxx}\color{blue}{\text{all terms to the left}}\\
(x+5)\cdot (x-3)&=&0 \\
&&\phantom{xxx}\color{blue}{\text{left hand side factored}}\\
x+5=0 &\lor& x-3=0\\
&&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\
x=-5 &\lor& x=3\\
&&\phantom{xxx}\color{blue}{\text{constant terms to the right}}
\end{array}\]

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