Finding minimum value of a square root function

Question

Given that x is a real number, find the minimum value of $\sqrt{(x+2017)(x+2018)(x+2019)(x+2020)+1}$

Attempt:

Let z= (x + 2018)

$\sqrt{(z-1)(z)(z+1)(z+2)+1}$

First derivative of the equation gives me a value of $z = -1/2$, giving a miniumum of 5/4?

But i tried plugging -2017 to the equation, giving a value of 1

Answer 1

You can just set it equal to $0$ and find the zeros, since it is a square root function, the values cannot be $<0$ in this case.

Here, you know it must have a real zero, since there is nothing in the function located "outside" the square root.

$\sqrt{(x+2017)(x+2018)(x+2019)(x+2020)+1}=0$

$(x+2017)(x+2018)(x+2019)(x+2020)+1=0$

Let $a=x+2017$

$a(a+1)(a+2)(a+3)+1=0$

$a^4+6a^3+11a^2+6a+1=0$

$(a^2 + 3a + 1)^2=0$

$a= \cfrac 32 \pm \cfrac{\sqrt5}{2}$

Since $a=x+2017$

$x+2017=\cfrac 32 \pm \cfrac{\sqrt5}{2}$

Answer 2

The main thing is $$ t(t+1)(t+2)(t+3) + 1 = \left( t^2 + 3 t + 1 \right)^2 $$ Let's see, we have $$ \left| t^2 + 3 t + 1 \right| $$ which does achieve zero when $$ t = \frac{-3 \pm \sqrt 5}{2} $$

Answer 3

I'm going to quibble that it asked what the minimum can be; not at which $x$ value it is achieved.

$\sqrt{(x+2017)(x+2018)(x+2019)(x+2020)+1} \ge 0$ so the absolute minimum possible is $0$, if $(x+2017)(x+2018)(x+2019)(x+2020) = -1$.

Let $a = x + 2018.5$ then $$(x+2017)(x+2018)(x+2019)(x+2020)= (a - \frac 32)(a-\frac 12)(a+\frac 12)(a+\frac 32) = (a^2 -\frac 94)(a^2 -\frac 14)=a^4 - \frac {10}4a^2 +\frac 9{16} = 1$$ if

$$a^2 = \frac {\frac {5}2 \pm \sqrt {\frac {25}4 - \frac9{4}}}{2} = \frac 54 \pm 2=\frac {13}4$$

So $\sqrt{(x+2017)(x+2018)(x+2019)(x+2020)+1} = 0$ if $x=\pm \frac{\sqrt{13}}2 + 2018.5$