Math Problem Statement

Solve for x in the triangle using the given expressions: (5x + 5), (15x - 3), and (7x).

Solution

The problem involves solving for xx using properties of a triangle. From the diagram, it appears we are working with a triangle and expressions along its sides. Let me extract the mathematical relationships to proceed.

  1. From the triangle, we observe three sides are labeled:
    • 5x+55x + 5 (shorter leg)
    • 15x315x - 3 (longer leg)
    • 7x7x (hypotenuse)

The triangle likely follows the Pythagorean theorem since it seems to be a right triangle: (shorter leg)2+(longer leg)2=(hypotenuse)2\text{(shorter leg)}^2 + \text{(longer leg)}^2 = \text{(hypotenuse)}^2 Substitute the values: (5x+5)2+(15x3)2=(7x)2(5x + 5)^2 + (15x - 3)^2 = (7x)^2

Expand and simplify

  1. Expand each term:

    • (5x+5)2=25x2+50x+25(5x + 5)^2 = 25x^2 + 50x + 25
    • (15x3)2=225x290x+9(15x - 3)^2 = 225x^2 - 90x + 9
    • (7x)2=49x2(7x)^2 = 49x^2
  2. Substitute into the equation: 25x2+50x+25+225x290x+9=49x225x^2 + 50x + 25 + 225x^2 - 90x + 9 = 49x^2

  3. Combine like terms: (25x2+225x249x2)+(50x90x)+(25+9)=0(25x^2 + 225x^2 - 49x^2) + (50x - 90x) + (25 + 9) = 0 201x240x+34=0201x^2 - 40x + 34 = 0

Solve the quadratic equation

Use the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=201a = 201, b=40b = -40, c=34c = 34.

x=(40)±(40)24(201)(34)2(201)x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(201)(34)}}{2(201)} x=40±160027336402x = \frac{40 \pm \sqrt{1600 - 27336}}{402} x=40±25736402x = \frac{40 \pm \sqrt{-25736}}{402}

Since the discriminant (25736-25736) is negative, there are no real solutions for xx.

Conclusion

This setup does not yield real solutions. The problem might need re-checking for alternative interpretations or if imaginary solutions are acceptable.

Would you like me to explain anything further or check another possible interpretation?


5 Related Questions:

  1. What is the significance of the Pythagorean theorem in geometry?
  2. How do we identify when a quadratic equation has no real solutions?
  3. Can this problem have solutions in the complex number system?
  4. Are there any specific conditions for right triangles that simplify such problems?
  5. What methods can help verify whether a solution to a geometry problem is plausible?

Tip: Always confirm that all triangle conditions are valid (e.g., sum of angles, side lengths) before solving equations involving geometric shapes.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations

Formulas

Pythagorean Theorem
Quadratic Formula

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-12