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To solve the problem, let's carefully analyze the given right triangle \( \triangle ABC \) where:

- \( \angle ACB = 90^\circ \) (indicating it's a right triangle)
- \( AB = 10x - 3 \) (hypotenuse)
- \( AC = 3x \) (one leg)
- \( BC = 5x + 2 \) (other leg)

We can use the Pythagorean theorem to find \( x \). The Pythagorean theorem states:

\[
AB^2 = AC^2 + BC^2
\]

### Step-by-Step Solution:

1. **Apply the Pythagorean theorem**:
   \[
   (10x - 3)^2 = (3x)^2 + (5x + 2)^2
   \]

2. **Expand and simplify each term**:
   \[
   (10x - 3)^2 = 100x^2 - 60x + 9
   \]
   \[
   (3x)^2 = 9x^2
   \]
   \[
   (5x + 2)^2 = 25x^2 + 20x + 4
   \]

3. **Combine and simplify the equation**:
   \[
   100x^2 - 60x + 9 = 9x^2 + 25x^2 + 20x + 4
   \]
   \[
   100x^2 - 60x + 9 = 34x^2 + 20x + 4
   \]

4. **Move all terms to one side**:
   \[
   100x^2 - 34x^2 - 60x - 20x + 9 - 4 = 0
   \]
   \[
   66x^2 - 80x + 5 = 0
   \]

5. **Solve the quadratic equation**:
   Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 66 \), \( b = -80 \), and \( c = 5 \).

   \[
   x = \frac{80 \pm \sqrt{(-80)^2 - 4(66)(5)}}{2(66)}
   \]
   \[
   x = \frac{80 \pm \sqrt{6400 - 1320}}{132}
   \]
   \[
   x = \frac{80 \pm \sqrt{5080}}{132}
   \]
   \[
   x = \frac{80 \pm 71.28}{132}
   \]

6. **Calculate the roots**:
   \[
   x_1 = \frac{80 + 71.28}{132} \approx \frac{151.28}{132} \approx 1.146
   \]
   \[
   x_2 = \frac{80 - 71.28}{132} \approx \frac{8.72}{132} \approx 0.066
   \]

### Choosing the correct root:
Since the problem states \( x > 0.5 \), the correct value of \( x \) is approximately **1.146**, but this is closer to **1.2**.

This doesn't align with the options provided, so there might be a rounding or simplification issue in the problem or available options.

However, if the problem is simplified, the closest answer could be **0.7** or **0.8**. For the purpose of this problem, given that we know \( x \) must be more than 0.5, the best available option could be **0.8** as it's closer to 1.146 than 0.7.

Would you like more details or have any questions about this solution?

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### Follow-up Questions:
1. What happens if you change the value of \( c \) in the quadratic equation?
2. How would the calculation change if \( x \) were less than 0.5?
3. What does the Pythagorean theorem tell us about the relationship between the sides of a right triangle?
4. How would the quadratic formula change if \( a \) were negative?
5. Can you estimate \( x \) using a different method, like completing the square?

### Tip:
Always check your quadratic solutions to ensure they make sense in the context of the problem, particularly when conditions like \( x > 0.5 \) are given.

Explore how to solve a right triangle problem using the Pythagorean theorem and quadratic equations. Step-by-step solution provided for finding the value of x in the given triangle. Suitable for Grades 9-10 students and those interested in geometry and algebra.

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