A right triangle has legs of lengths \( x \) and \( x+3 \), and a hypotenuse of length \( x+5 \). Find the value of \( x \).
["How to Solve for ( x ) in a Right Triangle Using the Pythagorean Theorem", "In geometry, right triangles provide a fundamental opportunity to apply the Pythagorean Theorem: ( a^2 + b^2 = c^2 ), where ( a ) and ( b ) are the legs, and ( c ) is the hypotenuse. Today, we explore a classic problem where a right triangle has legs of lengths ( x ) and ( x+3 ), and a hypotenuse of length ( x+5 ). Learning how to set up and solve this equation helps sharpen algebraic and mathematical reasoning—skills essential for students and math enthusiasts alike.", "---", "### The Problem: Right Triangle with Legs ( x ) and ( x+3 ), Hypotenuse ( x+5 )", "Given:\n- Leg 1: ( a = x )\n- Leg 2: ( b = x + 3 )\n- Hypotenuse: ( c = x + 5 )", "Apply the Pythagorean Theorem:\n[\nx^2 + (x + 3)^2 = (x + 5)^2\n]", "---", "### Step-by-step Solution", "1. Expand both sides of the equation:\n[\nx^2 + (x^2 + 6x + 9) = (x^2 + 10x + 25)\n]", "2. Simplify the left-hand side:\n[\nx^2 + x^2 + 6x + 9 = 2x^2 + 6x + 9\n]", "3. Set equation:\n[\n2x^2 + 6x + 9 = x^2 + 10x + 25\n]", "4. Move all terms to one side to form a quadratic equation:\nSubtract ( x^2 + 10x + 25 ) from both sides:\n[\n2x^2 + 6x + 9 - x^2 - 10x - 25 = 0\n]\n[\nx^2 - 4x - 16 = 0\n]", "---", "### Solving the Quadratic Equation", "We solve ( x^2 - 4x - 16 = 0 ) using the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nHere, ( a = 1 ), ( b = -4 ), ( c = -16 ).", "[\nx = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-16)}}{2(1)} = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2}\n]", "Simplify ( \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5} ):\n[\nx = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5}\n]", "---", "### Selecting the Valid Solution", "Since ( x ) represents a length, it must be positive:\n- ( x = 2 + 2\sqrt{5} ) (positive)\n- ( x = 2 - 2\sqrt{5} ) (negative, discard)", "---", "### Conclusion: The Value of ( x )", "The only valid solution is:\n[\nx = 2 + 2\sqrt{5}\n]", "This example illustrates how real-world geometric shapes translate into solvable algebraic equations—and how precision in calculation leads to accurate results. Whether you're solving for unknown side lengths or preparing for standardized tests, mastering the Pythagorean Theorem and quadratic reasoning is key.", "---", "Keywords: right triangle, Pythagorean Theorem, solve for ( x ), quadratic equation, algebra, geometry, math problem solving, educational example, ( x^2 - 4x - 16 = 0 ), right triangle with legs ( x ), hypotenuse ( x+5 )", "Meta Description:\nDiscover how to find ( x ) in a right triangle with legs ( x ) and ( x+3 ), and hypotenuse ( x+5 ) using the Pythagorean Theorem. Step-by-step solution with algebra and real math application. Perfect for students and math learners.", "Content Outline for SEO:\n- Title: How to Find ( x ) in a Right Triangle Using the Pythagorean Theorem\n- H2: Left Triangle Sides – Set Up the Equation\n- H2: Expand and Simplify: Solve for ( x )\n- H2: Final Quadratic and Solution\n- H2: Why This Matters – Applications in Geometry\n- H2: Summary and Common Mistakes", "Ready to Solve Your Next Geometry Problem?\nStart with the Pythagorean Theorem and algebraic accuracy—your next discovery awaits!"]
