Let's denote the distance traveled by the car heading east as $ x $ miles. According to the problem, the car traveling north has gone 24 miles. We need to find the distance between the two cars at that time.

Given that the distance between the cars is four miles more than three times the distance traveled by the car heading east, we can write this relationship as:
$$ \text{Distance between the cars} = 3x + 4 $$

Since the cars are traveling at right angles to each other, we can use the Pythagorean theorem to relate the distances. The distance between the cars forms the hypotenuse of a right triangle, with the legs being the distances traveled by each car. Therefore, we have:
$$ (\text{Distance between the cars})^2 = (\text{Distance traveled by the car heading north})^2 + (\text{Distance traveled by the car heading east})^2 $$

Substituting the given distances:
$$ (3x + 4)^2 = 24^2 + x^2 $$

Expanding and simplifying the equation:
$$ 9x^2 + 24x + 16 = 576 + x^2 $$

Rearranging the terms to form a quadratic equation:
$$ 9x^2 + 24x + 16 - x^2 - 576 = 0 $$
$$ 8x^2 + 24x - 560 = 0 $$

Dividing the entire equation by 8 to simplify:
$$ x^2 + 3x - 70 = 0 $$

Now, we solve this quadratic equation using the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 3 $, and $ c = -70 $:
$$ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-70)}}{2 \cdot 1} $$
$$ x = \frac{-3 \pm \sqrt{9 + 280}}{2} $$
$$ x = \frac{-3 \pm \sqrt{289}}{2} $$
$$ x = \frac{-3 \pm 17}{2} $$

This gives us two solutions:
$$ x = \frac{14}{2} = 7 $$
$$ x = \frac{-20}{2} = -10 $$

Since distance cannot be negative, we discard the negative solution. Therefore, the distance traveled by the car heading east is $ x = 7 $ miles.

Now, we can find the distance between the cars:
$$ \text{Distance between the cars} = 3x + 4 = 3(7) + 4 = 21 + 4 = 25 $$

Thus, the distance between the cars at that time is 25 miles.

Question

Let's denote the distance traveled by the car heading east as $ x $ miles. According to the problem, the car traveling north has gone 24 miles. We need to find the distance between the two cars at that time.

Given that the distance between the cars is four miles more than three times the distance traveled by the car heading east, we can write this relationship as:
$$ \text{Distance between the cars} = 3x + 4 $$

Since the cars are traveling at right angles to each other, we can use the Pythagorean theorem to relate the distances. The distance between the cars forms the hypotenuse of a right triangle, with the legs being the distances traveled by each car. Therefore, we have:
$$ (\text{Distance between the cars})^2 = (\text{Distance traveled by the car heading north})^2 + (\text{Distance traveled by the car heading east})^2 $$

Substituting the given distances:
$$ (3x + 4)^2 = 24^2 + x^2 $$

Expanding and simplifying the equation:
$$ 9x^2 + 24x + 16 = 576 + x^2 $$

Rearranging the terms to form a quadratic equation:
$$ 9x^2 + 24x + 16 - x^2 - 576 = 0 $$
$$ 8x^2 + 24x - 560 = 0 $$

Dividing the entire equation by 8 to simplify:
$$ x^2 + 3x - 70 = 0 $$

Now, we solve this quadratic equation using the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 3 $, and $ c = -70 $:
$$ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-70)}}{2 \cdot 1} $$
$$ x = \frac{-3 \pm \sqrt{9 + 280}}{2} $$
$$ x = \frac{-3 \pm \sqrt{289}}{2} $$
$$ x = \frac{-3 \pm 17}{2} $$

This gives us two solutions:
$$ x = \frac{14}{2} = 7 $$
$$ x = \frac{-20}{2} = -10 $$

Since distance cannot be negative, we discard the negative solution. Therefore, the distance traveled by the car heading east is $ x = 7 $ miles.

Now, we can find the distance between the cars:
$$ \text{Distance between the cars} = 3x + 4 = 3(7) + 4 = 21 + 4 = 25 $$

Thus, the distance between the cars at that time is 25 miles.

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