# find three consecutive integers such that the sum of the squares of the second and the third numbers exceeds the square of the first number by 21

2
by ayalyn

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by ayalyn

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x+1 be the 2nd number

x+2 be the 3rd number

as stated in the problem, we can form the equation:

(x+1)² + (x+2)² = x² + 21

expanding the terms we'll have

(x² + 2x + 1) + (x² + 4x +4) = x² + 21

combine the like terms and transposing x² + 21 to the left:

x²+x²-x² + 2x + 4x +1 + 4 - 21 = 0

x² + 6x -16 = 0

factor:

(x + 8) (x - 2) = 0

getting the roots:

x + 8 = 0

x = - 8

x - 2 = 0

x = 2

taking the first value of x which is -8,

the consecutive negative integers would be:

x = -8

x + 1 = -7

x + 2= -6

(-8,-7,-6)

considering the second value of x which is 2

the consecutive positive integers would be:

x= 2

x + 1 = 3

x + 2 = 4

(2,3,4)

[(x+1)^2+(x+2)^2)]-x^2=21

[(x^2+2x+1)+(x^2+4x+4)]-x^2=21

(2x^2+6x+5)-x^2=21

x^2+6x+5=21

x^2+6x-16=0

(x+8)(x-2)=0

x=-8; x=2

the numbers are -8, -7, -6 or 2, 3, 4