Possible x-Values from a 40-Unit Distance

Problem

The distance between the points $(6, 7)$ and $(x, -25)$ is $40$ units. Find the possible values of $x$.

Step 1: Identify the coordinate changes

The two points have $x$-coordinates $6$ and $x$, and $y$-coordinates $7$ and $-25$.

The horizontal change is

$$x - 6$$

and the vertical change is

$$-25 - 7 = -32.$$

Step 2: Set up the distance formula

Using the distance formula,

$$d = \sqrt{(\text{horizontal change})^2 + (\text{vertical change})^2},$$

and substituting the given distance, we get

$$40 = \sqrt{(x - 6)^2 + (-32)^2}.$$

Step 3: Remove the square root

Square both sides to remove the square root:

$$40^2 = (x - 6)^2 + (-32)^2.$$

So,

$$1600 = (x - 6)^2 + 1024.$$

Step 4: Isolate the squared term

Subtract $1024$ from both sides:

$$1600 - 1024 = (x - 6)^2.$$

Thus,

$$(x - 6)^2 = 576.$$

Step 5: Solve both square-root cases

Since

$$(x - 6)^2 = 576,$$

we have

$$x - 6 = 24$$

or

$$x - 6 = -24.$$

Solving each equation gives

$$x = 30$$

or

$$x = -18.$$

Step 6: State the possible values

Both values work because

$$24^2 + 32^2 = 40^2.$$

Therefore, the possible values of $x$ are

$$\boxed{30 \text{ and } -18}.$$