Answer
$\overline{AB}$ is $6$.
Work Step by Step
We are given that $\overline{AM}$ is $x^2 - 6$ and $\overline{MB}$ is $x$.
If $M$ is the midpoint of $\overline{AB}$, then $\overline{AM}$ would equal $\overline{MB}$; this means that we can set them equal to one another:
$x^2 - 6 = x$
It looks like we have a quadratic equation, which is given by the formula $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers.
Let's subtract $x$ from both sides of the equation to set the quadratic equation equal to $0$:
$x^2 - x - 6 = 0$
We can now factor this equation. We need to find the factors of $ac$ or $-6$ that, when added together, will equal $b$ or $-1$. We need one negative factor and one positive factor because a positive times a negative will equal a negative; when they are added together, either a negative or a positive number can result, depending on the absolute value of the negative and positive number. In this case, we need the negative number to have the greater absolute value.
We found the following possibilities:
$(a)(c) = (-6)(1)$
$b = -5$
$(a)(c) = (-3)(2)$
$b = -1$
The second option works for us. Now we can factor the quadratic equation:
$(x - 3)(x + 1) = 0$
According to the zero product property, if the product of two factors is $0$, then each factor is equal to $0$ or both factors equal $0$. Let us now set each factor equal to zero:
First factor:
$x - 3 = 0$
Add $3$ to each side to solve for $x$:
$x = 3$
Second factor:
$x + 1 = 0$
Subtract $1$ from each side to solve for $x$:
$x = -1$
We can only use the first solution $x = 3$ because we cannot have a negative distance.
So if $\overline{MB}$ is $3$, then $\overline{AM}$ is also $3$. This means that $\overline{AB}$ is $6$.
