By completing the square, we have
[tex](x+1)(x+4) = x^2 + 5x + 4 = \left(x+\dfrac52\right)^2 - \dfrac94[/tex]
[tex](x+2)(x+3) = x^2 + 5x + 6 = \left(x + \dfrac52\right)^2 - \dfrac14[/tex]
Then
[tex](x+1)(x+2)(x+3)(x+4) = \left(x+\dfrac52\right)^4 - \dfrac52 \underbrace{\left(x + \dfrac52\right)^2}_{=y} + \dfrac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = y^2 - \dfrac52 y + \dfrac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(y - \dfrac54\right)^2 - 1 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(\left(x + \dfrac52\right)^2 - \dfrac54\right)^2 - 1[/tex]
This product has a minimum of -1 when
[tex]\left(x + \dfrac52\right)^2 - \dfrac54 = 0[/tex]
which has two real solutions; then the minimum of the overall expression is -1 + 2019 = 2018.
The least possible value of (x+1)(x+2)(x+3)(x+4)+2019 where x is a real number is 2018
An expression is a mathematical statement consisting of variables , constant and mathematical operators.
The expression given is
(x+1)(x+2)(x+3)(x+4)+2019
To find the minimum value
(x+1)(x+2) = ( x² +3x +2) = ( x +5/2)² - 9/4
(x+3)(x+4) = (x² +7x +12) = ( x +5/2)² -1/4
(x+1)(x+2)(x+3)(x+4) = ( x +5/2)⁴ - 5/2 ( x +5/2)² +9/16
Considering ( x +5/2)² = y
(x+1)(x+2)(x+3)(x+4) = y² - 5/2y + 9/16
(x+1)(x+2)(x+3)(x+4) = ( y - 5/4)² - 1
(x+1)(x+2)(x+3)(x+4) = ( ( x +5/2)² -5/4)² - 1
The minimum value when x is a real number is -1
( x +5/2)² -5/4) = 0
-1 +2019 = 2018
x has two possible real solution
and the least possible value of (x+1)(x+2)(x+3)(x+4)+2019 where x is a real number is 2018.
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