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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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
Darrell
4.5100 Answers
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: xΒ² + (x + 8)Β² = 80
**2. Solve the equation:**
* Expand the equation: xΒ² + xΒ² + 16x + 64 = 80
* Combine like terms: 2xΒ² + 16x - 16 = 0
* Divide by 2 for simplicity: xΒ² + 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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