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quadratic equations. A positive real number is 4 more than another. If the sum of the squares of the two numbers is 56, find the numbers.

User Dobbs
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1 Answer

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Answer:

2√6 +2 and 2√6 -2

Explanation:

You want two numbers such that one is 4 more than the other and the sum of their squares is 56.

Setup

Let x represent the smaller number. The the sum of the squares of the two numbers is ...

x² +(x +4)² = 56

Solution

Simplifying the equation, we have ...

2x² +8x +16 = 56

x² +4x = 20 . . . . . . . . . subtract 16, divide by 2

x² +4x +4 = 24 . . . . . . add 4 to complete the square

(x +2)² = 2²·6 . . . . . . . write in terms of squares

x = 2√6 -2 . . . . . . . . positive square root, subtract 6

The two numbers are 2√6 -2 and 2√6 +2.

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Additional comment

Their approximate decimal values are 6.8990 and 2.8990. As you expect, the sum of the squares of the rounded values differs slightly from 56.

quadratic equations. A positive real number is 4 more than another. If the sum of-example-1
User Bamnet
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