Solving a simple quadratic equation : $40x+3 = (x+14)(x+3)+(x+1)$

Question

I have the equation from one of the textbooks (which is, I know, very simple):

$40x+3 = (x+14)(x+3)+(x+1)$. My first approach to the solution is as follows

$$\begin{align*} 40x+3&=x^{2} +14x +3x +42 +x +1 \\[0.1cm] 40x+3 &= x^{2} +18x +43\\[0.1cm] 40x &= x^{2} +18x + 40 \\[0.1cm] 0 &= x^{2} -22x +40 \\[0.1cm] 0 &= (x-2)(x-20) \end{align*}$$

and it follows that $x_{1} = 2$ and $x_{2} = 20$ (these are also the solutions in the textbook). The second approach:

$$\begin{align*} 40x+3&=x^{2} +14x +3x +42 +x +1 \\[0.1cm] 40x+3 &= x^{2} +18x +43\\[0.1cm] 40x &= x^{2} +18x +40 \\[0.1cm] 22x &= x^{2} + 40 \\[0.1cm] -x^{2} +22x -40 &= 0 \\[0.1cm] - (x-2)(x-20)&= 0 \end{align*}$$

Why do I get $- (x-2)(x-20)$ with the second approach, I should get the same answer as before. Where is the mistake?

Answer 1

Negatives can trip people up easily at first:

You have reduced things to $0 = x^{2} -22x + 40 = (x-2)(x-20)$

Note that by subtracting from both sides:

$(x-2)(x-20)= 0$

$(x-2)(x-20)-(x-2)(x-20) = 0 -(x-2)(x-20)$

$ \Rightarrow 0=-(x-2)(x-20)$

So it is equivalent to solve $0=-(x-2)(x-20)$ and $0=(x-2)(x-20)$.

This kind of situation arises often, you just have to get used to it.

Further, note:

$(x-2)(x-20)= 0$

$(-1)(2-x)(-1)(20-x)= 0$

$(2-x)(20-x)= 0$

So it is equivalent to solve $0=(x-2)(x-20)$ and $0=(2-x)(20-x)$.