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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

Sagot :

Shinalcantaraidn
let x be the first number
    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)
Rmmerencillaidn
(x, x+1, x+2)
[(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

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