Question

What is the least possible value of (x+1)(x+2)(x+3)(x+4)+2019 where x is a real number?

Answer 1

By completing the square, we have

$(x+1)(x+4) = x^2 + 5x + 4 = \left(x+\dfrac52\right)^2 - \dfrac94$

$(x+2)(x+3) = x^2 + 5x + 6 = \left(x + \dfrac52\right)^2 - \dfrac14$

Then

$(x+1)(x+2)(x+3)(x+4) = \left(x+\dfrac52\right)^4 - \dfrac52 \underbrace{\left(x + \dfrac52\right)^2}_{=y} + \dfrac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = y^2 - \dfrac52 y + \dfrac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(y - \dfrac54\right)^2 - 1 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(\left(x + \dfrac52\right)^2 - \dfrac54\right)^2 - 1$

This product has a minimum of -1 when

$\left(x + \dfrac52\right)^2 - \dfrac54 = 0$

which has two real solutions; then the minimum of the overall expression is -1 + 2019 = 2018.

Answer 2

The least possible value of (x+1)(x+2)(x+3)(x+4)+2019 where x is a real number is 2018

What is an Expression ?

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.

To know more about Expression

brainly.com/question/14083225

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