Find four consecutive integers such that the sum of the squares of the first two is 11 less than the square of the fourth.

Anish Buchanan

Anish Buchanan

Answered question

2020-12-01

Find four consecutive integers such that the sum of the squares of the first two is 11 less than the square of the fourth.

Answer & Explanation

liingliing8

liingliing8

Skilled2020-12-02Added 95 answers

Consecutive integers differ by 1 so we let the four consecutive integers be:
x,x+1,x+2,x+3
The sum of the squares of the first two is 11 less than the square of the fourth so:
x2+(x+1)2=(x+3)211
Solve for x:
x2+(x2+2x+1)=(x2+6x+9)11
2x2+2x+1=x2+6x2
x24x+3=0
Factor the left side: (x−1)(x−3)=0
By zero product property, x=1,3
There are two possible sets of consecutive integers: 1,2,3,4 or 3,4,5,6

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