Math Problem Statement

Find the value of x:

Solution

To find the value of xx in the given geometric configuration, we can use the properties of right triangles and the Pythagorean theorem.

Given:

  • The right triangle with legs xx and x+5x + 5
  • The hypotenuse of this right triangle is 20

Using the Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2

Here, a=xa = x, b=x+5b = x + 5, and c=20c = 20. Thus, we have: x2+(x+5)2=202x^2 + (x + 5)^2 = 20^2

Expanding and simplifying the equation: x2+(x2+10x+25)=400x^2 + (x^2 + 10x + 25) = 400 2x2+10x+25=4002x^2 + 10x + 25 = 400 2x2+10x+25400=02x^2 + 10x + 25 - 400 = 0 2x2+10x375=02x^2 + 10x - 375 = 0

Dividing the entire equation by 2 to simplify: x2+5x187.5=0x^2 + 5x - 187.5 = 0

This is a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. We can solve it using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation, a=1a = 1, b=5b = 5, and c=187.5c = -187.5: x=5±5241(187.5)21x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-187.5)}}{2 \cdot 1} x=5±25+7502x = \frac{-5 \pm \sqrt{25 + 750}}{2} x=5±7752x = \frac{-5 \pm \sqrt{775}}{2} x=5±25312x = \frac{-5 \pm \sqrt{25 \cdot 31}}{2} x=5±5312x = \frac{-5 \pm 5\sqrt{31}}{2} x=5+5312orx=55312x = \frac{-5 + 5\sqrt{31}}{2} \quad \text{or} \quad x = \frac{-5 - 5\sqrt{31}}{2}

Since xx represents a length, we take the positive value: x=5+5312x = \frac{-5 + 5\sqrt{31}}{2}

Simplifying further: x=5(311)2x = \frac{5(\sqrt{31} - 1)}{2}

Thus, the value of xx is: x6.1x \approx 6.1

Would you like further details or have any questions?

Here are five more questions you can ask:

  1. How do I solve another quadratic equation?
  2. Can you explain the steps of the quadratic formula?
  3. What are the properties of right triangles?
  4. How is the Pythagorean theorem used in different types of problems?
  5. How can I verify the solution to a geometric problem?

Tip: Always double-check your algebraic manipulations when solving equations to avoid small errors that can lead to incorrect solutions.

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Math Problem Analysis

Mathematical Concepts

Geometry
Pythagorean Theorem
Quadratic Equations

Formulas

Pythagorean theorem
Quadratic formula

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 9-12