Solving equations with factorization

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

Consider the quadratic equation with unknown \(x\):

\(ax^2+bx+c=0\),

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 \(x=p\lor x=q\).


Let’s practice!

Solve the equation for \(x\) below using factorization.

\(x^2-14 x-16=-8 x\tiny.\)

Give your answer in the form of \(x=a\lor x=b\).

This can be solved by means of the following deduction:

\(\begin{array}{rcl} x^2-14 x-16&=&-8 x \\ &&\phantom{xxx}\color{blue}{\text{the original equation}}\\ x^2-6 x-16&=&0 \\ &&\phantom{xxx}\color{blue}{\text{all terms to the left}}\\ (x-8)\cdot (x+2)&=&0 \\ &&\phantom{xxx}\color{blue}{\text{left hand side factored}}\\ x-8=0 &\lor& x+2=0\\ &&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ x=8 &\lor& x=-2\\ &&\phantom{xxx}\color{blue}{\text{constant terms to the right}} \end{array}\)

Answer:

\(x=8\lor x=-2\)

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