find x so that the distance between the points (x,1) and (3,4) is $$ \sqrt{34} $$

We are given two points: $ P(x, 1) $ and $ Q(3, 4) $. The distance between these two points is $ \sqrt{34} $. 

The distance formula between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

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

Using this formula, the distance between $ P(x, 1) $ and $ Q(3, 4) $ is:

\[
\sqrt{(3 - x)^2 + (4 - 1)^2} = \sqrt{34}
\]

Simplifying the equation inside the square root:

\[
\sqrt{(3 - x)^2 + 3^2} = \sqrt{34}
\]

\[
\sqrt{(3 - x)^2 + 9} = \sqrt{34}
\]

Now, square both sides to eliminate the square roots:

\[
(3 - x)^2 + 9 = 34
\]

Subtract 9 from both sides:

\[
(3 - x)^2 = 25
\]

Now, take the square root of both sides:

\[
3 - x = \pm 5
\]

So, we have two possible cases:

1. $ 3 - x = 5 $
2. $ 3 - x = -5 $

### Case 1: $ 3 - x = 5 $

\[
x = 3 - 5 = -2
\]

### Case 2: $ 3 - x = -5 $

\[
x = 3 + 5 = 8
\]

Thus, the two possible values of $ x $ are $ x = -2 $ and $ x = 8 $.

### Conclusion:
The values of $ x $ that satisfy the given condition are $ x = -2 $ and $ x = 8 $.

Would you like more details, or do you have any questions?

Here are five related questions:
1. How do you find the distance between two points in three-dimensional space?
2. Can there be more than two solutions for distance problems like this?
3. How would the result change if the distance given was a different value?
4. What is the general equation for a circle given a fixed distance from a point?
5. How can the distance formula be derived from the Pythagorean theorem?

**Tip**: Always remember to check both positive and negative square root solutions when solving quadratic equations!

Find x so that the distance between the points (x, 1) and (3, 4) is sqrt(34).

Learn how to find the value of x so that the distance between points (x, 1) and (3, 4) equals sqrt(34). This solution covers the distance formula and solving quadratic equations. Suitable for students in Grades 9-11.

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