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bezimeni [28]
4 years ago
10

Two consecutive positive numbers are such that the sum of their squares is 113. find the two numbers

Mathematics
1 answer:
Naya [18.7K]4 years ago
8 0

Answer: 7 and 8

<u>Step-by-step explanation:</u>

Let x represent the first number, then x + 1 is the other number.

(x)² + (x + 1)² = 113

x²  +  x² + 2x + 1 = 113    <em>expanded (x + 1)²</em>

2x² + 2x + 1 = 113          <em>added like terms</em>

2x² + 2x - 112 = 0        <em>subtracted 113 from both sides</em>

x² + x - 56 = 0 <em>           divided both sides by 2</em>

(x + 8) (x - 7) = 0        <em>factored polynomial</em>

x + 8 = 0   x - 7 = 0    <em>applied zero product property</em>

x = -8     x = 7  <em>          solved for x</em>

  ↓

not valid since the restriction is that x > 0 <em>(a positive number)</em>

So, x = 7     and     x + 1 = (7) + 1   = 8

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Then insert the assigned values into this equation:  \frac{y2-y1}{x2-x1} -> \frac{9-5}{-8-10} which equals \frac{4}{-18} which simplifies to -\frac{2}{9}

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