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The product of three consecutive integers x-1, x, and x+1 is 990. Find the numbers.

Sagot :

Solution:
x³ + x² - x² - x = 990
x³ - x = 990
x³ - x - 990 = 0

Factor:
(x-10) (x² + 10x  + 99) = 0

x - 10 = 0
x = 10

Solve  x² +10x + 99 = 0 by Completing the Square or Quadratic Formula.
By Completing the Square:
ax² + bx + c = 0
ax² + bx + (b/2)² = -c + (b/2)²

b = 10
x² + 10x + (10/2)² = -99 + (10/2)²
x² + 10x + (100/4) = -99 + (100/4)
x² + 10x + 25 = -99 + 25
(x + 5) (x + 5) = -74
(x + 5)² = -74
[tex] \sqrt{(x+5) ^{2} } = \sqrt{-74} [/tex]
[tex]x + 5 = \sqrt{-74} [/tex]

[tex] \sqrt{-74} [/tex] is an imaginary number:
[tex] \sqrt{-74} = +or-i \sqrt{74} [/tex]
[tex]x + 5 -5 = -5 +i \sqrt{74} [/tex]
      and
[tex]x+5-5 =-5-i \sqrt{74} [/tex]

Therefore the roots (x) are:

x = 10

[tex]x = -5+i \sqrt{74} [/tex]

[tex]x = -5-i \sqrt{74} [/tex]