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a positive real number is 8 more than another if the sum of the squares of the two numbers is 80 find the numbers

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A positive real number is 8 more than another. If the sum of the squares of the two numbers is 80, find the numbers

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Answer to a math question A positive real number is 8 more than another. If the sum of the squares of the two numbers is 80, find the numbers

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Absolutely, let's solve this problem. Here's how to find the numbers: **1. Set up the equations:** * Let 'x' be the first number. * The second number is 'x + 8' (since it's 8 more than the first). * The sum of their squares is 80: x² + (x + 8)² = 80 **2. Solve the equation:** * Expand the equation: x² + x² + 16x + 64 = 80 * Combine like terms: 2x² + 16x - 16 = 0 * Divide by 2 for simplicity: x² + 8x - 8 = 0 * Factor the quadratic equation: (x + 4)(x - 2) = 0 * Solve for possible values of x: * x + 4 = 0 => x = -4 (We discard this since we need a positive real number) * x - 2 = 0 => x = 2 **3. Find the second number:** * Since x = 2, the second number is x + 8 = 2 + 8 = 10 **Answer:** The two numbers are 2 and 10.

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