Find x so that the distance between x,3 and 2,-1 is 5

To find \( x \) so that the distance between the points \( (x, 3) \) and \( (2, -1) \) is 5, we can use the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Here, we have:
- \( (x_1, y_1) = (x, 3) \)
- \( (x_2, y_2) = (2, -1) \)
- \( d = 5 \)

Plugging in these values, we get:

\[
5 = \sqrt{(2 - x)^2 + (-1 - 3)^2}
\]

Simplify the expression inside the square root:

\[
5 = \sqrt{(2 - x)^2 + (-4)^2}
\]
\[
5 = \sqrt{(2 - x)^2 + 16}
\]

Now, square both sides to remove the square root:

\[
25 = (2 - x)^2 + 16
\]

Subtract 16 from both sides:

\[
9 = (2 - x)^2
\]

Now, take the square root of both sides, remembering to include both the positive and negative solutions:

\[
2 - x = \pm 3
\]

This gives us two equations to solve:

1. \( 2 - x = 3 \)
2. \( 2 - x = -3 \)

#### Solve the first equation:
\[
2 - x = 3
\]
\[
-x = 1
\]
\[
x = -1
\]

#### Solve the second equation:
\[
2 - x = -3
\]
\[
-x = -5
\]
\[
x = 5
\]

So, the values of \( x \) that make the distance between the points equal to 5 are:
\[
x = -1 \quad \text{or} \quad x = 5
\]

Would you like additional details or explanations? 

Here are 5 related questions:
1. How would you find \( x \) if the distance was 10 instead of 5?
2. Can you determine the midpoint of points \( (x, 3) \) and \( (2, -1) \)?
3. What is the distance formula for points in 3D space?
4. How do we handle complex solutions if they arise in distance problems?
5. How can the distance formula be applied to find the radius of a circle centered at \( (2, -1) \)?

**Tip:** Always check both positive and negative solutions when solving squared equations, as they can yield multiple values.

Find x so that the distance between (x, 3) and (2, -1) is 5

Learn how to solve for x in the distance formula problem where the distance between (x, 3) and (2, -1) equals 5. This step-by-step guide covers the use of the distance formula, square roots, and solving quadratic equations. Ideal for students in Grades 8-10.

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