Find three consecutive integers such that the square of the sum of the first two is 80 more than the sum of the squares of the last two.
Answer to Problem 10MPS
7,8,9
-6,-5,-4
Explanation of Solution
Calculation:
Let the three consecutive numbers be :
x − 1 , x , x +1
Sum of the first two numbers = ( x − 1 ) + x = 2x − 1
Square of the sum of the first two numbers =
Sum of the squares of the last two numbers =
So,
Case I : x = -5
So, the numbers are -6 , -5 , -4
Case II : x = 8
So, the numbers are 7 , 8 , 9
Chapter 10 Solutions
Algebra and Trigonometry: Structure and Method, Book 2
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