At what value of x is the distance between points C (3, 2) and D (x; -1) equal to 5?

The distance between points A (x1; y1) and B (x2; y2) is calculated by the formula:
AB = √ ((x1 – x2) ^ 2 + (y1 – y2) ^ 2).
Substitute the known values:
CD = √ ((3 – x) ^ 2 + (2 – (-1)) ^ 2).
Since CD = 5, then:
√ ((3 – x) ^ 2 + (2 – (-1)) ^ 2) = 5.
Let’s solve this equation with one unknown. Let’s expand the brackets:
√ (3 ^ 2 – 2 * 3 * x + x ^ 2 + (2 + 1) ^ 2) = 5;
√ (9 – 6x + x ^ 2 + 3 ^ 2) = 5;
√ (9 – 6x + x ^ 2 + 9) = 5;
√ (x ^ 2 – 6x + 18) = 5.
Let’s square both sides of the equation to get rid of irrationality:
x ^ 2 – 6x + 18 = 25;
x ^ 2 – 6x + 18 – 25 = 0;
x ^ 2 – 6x – 7 = 0.
Discriminant:
D = b ^ 2 – 4ac = (-6) ^ 2 – 4 * 1 * (- 7) = 36 + 28 = 64.
x = (-b +/- √D) / 2a.
x1 = (- (- 6) + √64) / 2 * 1 = (6 + 8) / 2 = 14/2 = 7.
x2 = (- (- 6) – √64) / 2 * 1 = (6 – 8) / 2 = -2/2 = -1.
Answer: x1 = 7; x2 = -1.