Question

Solve the equation

\frac{ (x^2+3x+2)(x^2-6x+8) }{ x^2-4 }=3x-12

When I go to solve it, I end up with
(x+1)(x-4)=3x-12
(x-4)(x-2)=0

And x=4,2

BUT I know 2 is not a valid solution. How does this question work algebraically?

Answer 1

\[\frac{ (x^2+3x+2)(x^2-6x+8) }{ x^2-4 }=3x-12\]

Answer 2

\(\color{purple}{\displaystyle\frac{ (x^2+3x+2)(x^2-6x+8) }{ x^2-4 }=3x-12 }\)\(\tiny \\[0.6em]\)
------------------------------------------\(\tiny \\[0.6em]\)
Factor: \(\color{purple}{\displaystyle\frac{ (x+1)(x+2)(x-2)(x-4) }{ (x-2)(x+2) }=3x-12 }\)

your denominator can't be equal to 0, so assume \(\color{black}{\displaystyle x\ne\pm2}\).

Then, since \(\color{black}{\displaystyle x\ne\pm2}\), you can divide top and bottom by \(\color{black}{\displaystyle (x-2)(x+2)}\), (because \(\color{black}{\displaystyle (x-2)(x+2)\ne0}\), so you wouldn't be dividing by 0).

So you get: \(\color{purple}{\displaystyle (x+1)(x-4)=3x-12 }\)
\(\color{purple}{\displaystyle (x+1)(x-4)=3(x-4) }\)

Notice that if \(\color{black}{\displaystyle x-4=0 }\) your equation is satisfied,
so \(\color{black}{\displaystyle x=4 }\) is therefore a solution (solution #1).

Answer 3

So, having excluded the solution \(\color{black}{\displaystyle x=4 }\), we will assume \(\color{black}{\displaystyle x\ne4 }\), and then division by \(\color{black}{\displaystyle x-4 }\) on both sides is valid. This division gives you,

\(\color{black}{\displaystyle x+1=3 }\)
\(\color{black}{\displaystyle x=2 }\)

Answer 4

But, since \(\color{black}{x=2}\) has been already excluded (that is your restriction for the denominator \(\color{black}{x^2-4}\) not to be \(\color{black}{0}\)), therefore the only working solution is \(\color{black}{x=4}\).