Question

$\frac{x^2 - 5x + 6}{x - 2}$ and test the solution(s) by analyzing the graph of the equation shown below. what is the extraneous solution?$\bigcirc$ $x=4LXB0\bigcirc$ $x=3$

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation:
$$\frac{x^2 - 5x + 6}{x - 2} = 0$$
Factor the numerator:
$$\frac{(x-2)(x-3)}{x - 2} = 0$$

Step2: Simplify (exclude undefined point)

Cancel $x-2$ (note $x
eq2$, since this makes denominator 0):
$$x - 3 = 0 \implies x=3$$

Step3: Identify extraneous solution

The original equation is undefined at $x=2$, so $x=2$ is not a solution. From the graph, the vertical asymptote is at $x=2$, and the graph intersects the x-axis at $x=3$. The option $x=4$ does not satisfy the equation, but wait—correction: the extraneous solution is the value that makes the denominator zero but was incorrectly included if we didn't check. Wait, no: solving $\frac{(x-2)(x-3)}{x-2}=0$, if we don't exclude $x=2$, we might think $x=2$ is a solution, but it's not. But the options are $x=4, x=2, x=3$. The valid solution is $x=3$, so the extraneous solution is $\boldsymbol{x=2}$? Wait no, recheck:
Wait the equation is $\frac{x^2-5x+6}{x-2}=0$. The numerator is zero at $x=2$ and $x=3$, but $x=2$ makes denominator zero, so it's not a valid solution. So $x=2$ is extraneous. But wait the graph: the red/blue curves have a hole at $x=2$, asymptote? No, the graph shows that at $x=2$, there's a discontinuity. The solution to the equation is $x=3$, so the extraneous solution is $x=2$. Wait, but let's test:
For $x=2$: $\frac{4-10+6}{0}=\frac{0}{0}$ undefined, not a solution.
For $x=3$: $\frac{9-15+6}{1}=0$, valid.
For $x=4$: $\frac{16-20+6}{2}=\frac{2}{2}=1
eq0$, not a solution.
So the extraneous solution (the value that appears to be a solution but is not) is $\boldsymbol{x=2}$.

Wait, correction to steps:

Step1: Factor the numerator

$$x^2-5x+6=(x-2)(x-3)$$

Step2: Set equation to zero

$$\frac{(x-2)(x-3)}{x-2}=0$$

Step3: Exclude undefined value

$x
eq2$ (denominator cannot be 0)

Step4: Solve simplified equation

$x-3=0 \implies x=3$

Step5: Identify extraneous value

$x=2$ is the value that would seem to be a solution (makes numerator 0) but is undefined, so it's extraneous.

Step1: Factor the numerator

$x^2-5x+6=(x-2)(x-3)$

Step2: Write factored equation

$$\frac{(x-2)(x-3)}{x-2}=0$$

Step3: Exclude undefined $x$

$x
eq2$ (denominator = 0 here)

Step4: Solve simplified equation

$x-3=0 \implies x=3$

Step5: Identify extraneous solution

$x=2$ is invalid, so it is extraneous.

Answer:

$\boldsymbol{x=4}$