# Consider an equation log_2 (alpha^2-16alpha^3 +66) + sqrt(4beta^4 -8beta^2 +13)+ | (gamma/3-2)| = 4, Find the numbers of ordered triplets (alpha, beta, gamma)? Also find the sum of all possible values of the product alpha beta gamma?

Mar 20, 2017

There are two solutions:

$\left(\alpha , \beta , \gamma\right) = \left(2 , \pm 1 , 6\right)$

Hence the sum of possible values of $\alpha \beta \gamma$ is $0$.

#### Explanation:

The question should have had ${\alpha}^{6}$ instead of ${\alpha}^{2}$ (checked against original question sheet).

Given:

${\log}_{2} \left({\alpha}^{6} - 16 {\alpha}^{3} + 66\right) + \sqrt{4 {\beta}^{4} - 8 {\beta}^{2} + 13} + \left\mid \frac{\gamma}{3} - 2 \right\mid = 4$

Let us look at each subexpression in turn:

$\textcolor{w h i t e}{}$
($\boldsymbol{\alpha}$):

${\alpha}^{6} - 16 {\alpha}^{3} + 66 = {\left({\alpha}^{3}\right)}^{2} - 16 \left({\alpha}^{3}\right) + 64 + 2$

$\textcolor{w h i t e}{{\alpha}^{6} - 16 {\alpha}^{3} + 66} = {\left({\alpha}^{3} - 8\right)}^{2} + 2$

$\textcolor{w h i t e}{{\alpha}^{6} - 16 {\alpha}^{3} + 66} \ge 2$

taking the minimum value $2$ only when ${\alpha}^{3} = 8$, that is when $\alpha = 2$.

So:

${\log}_{2} \left({\alpha}^{6} - 16 {\alpha}^{3} + 66\right) \ge {\log}_{2} 2 = 1$

only taking the minimum value $1$ when $\alpha = 2$.

$\textcolor{w h i t e}{}$
($\boldsymbol{\beta}$):

$4 {\beta}^{2} - 8 {\beta}^{2} + 13 = 4 {\beta}^{2} - 8 {\beta}^{2} + 4 + 9$

$\textcolor{w h i t e}{4 {\beta}^{2} - 8 {\beta}^{2} + 13} = 4 \left({\left({\beta}^{2}\right)}^{2} - 2 {\beta}^{2} + 1\right) + 9$

$\textcolor{w h i t e}{4 {\beta}^{2} - 8 {\beta}^{2} + 13} = 4 {\left({\beta}^{2} - 1\right)}^{2} + 9$

taking the minimum value $9$ when ${\beta}^{2} = 1$, i.e. when $\beta = \pm 1$.

Hence:

$\sqrt{4 {\beta}^{2} - 8 {\beta}^{2} + 13} \ge \sqrt{9} = 3$

taking the minimum value $3$ when $\beta = \pm 1$.

$\textcolor{w h i t e}{}$
($\boldsymbol{\gamma}$):

$\left\mid \frac{\gamma}{3} - 2 \right\mid$

takes its minimum possible value $0$ when:

$\frac{\gamma}{3} - 2 = 0$

That is, when $\gamma = 6$

$\textcolor{w h i t e}{}$
Sum:

So the minimum possible value of:

${\log}_{2} \left({\alpha}^{6} - 16 {\alpha}^{3} + 66\right) + \sqrt{4 {\beta}^{4} - 8 {\beta}^{2} + 13} + \left\mid \frac{\gamma}{3} - 2 \right\mid$

is $1 + 3 + 0 = 4$, only occuring when $\alpha = 2$, $\beta = \pm 1$ and $\gamma = 6$.

So the only possible solutions of the original equation are:

$\left(\alpha , \beta , \gamma\right) = \left(2 , \pm 1 , 6\right)$

Hence, the sum of all possible values of $\alpha \beta \gamma$ is:

$\left(2 \cdot 1 \cdot 6\right) + \left(2 \cdot \left(- 1\right) \cdot 6\right) = 12 - 12 = 0$

Mar 20, 2017

See below.

#### Explanation:

We have a relationship as

$f \left(\alpha\right) + g \left(\beta\right) + p \left(\gamma\right) = 4$

with

%%%%%%%%%%%%%%%%%%%
$f \left(\alpha\right) = {\log}_{2} \left({\alpha}^{2} - 16 {\alpha}^{3} + 66\right)$
$g \left(\beta\right) = \sqrt{4 {\beta}^{4} - 8 {\beta}^{2} + 13}$
$p \left(\gamma\right) = \left\mid \frac{\gamma}{3} - 2 \right\mid$
%%%%%%%%%%%%%%%%%%%

Considering

$f \left(\alpha\right) = {\log}_{2} \left({\alpha}^{2} - 16 {\alpha}^{3} + 66\right)$

the conditions on $f \left(\alpha\right)$ are:

${\alpha}^{2} - 16 {\alpha}^{3} + 66 > 0$
$0 \le f \left(\alpha\right) \le 4$

or

$1.48314 \le \alpha < 1.62487$

The conditions for $g \left(\beta\right)$ are:

$4 {b}^{4} - 8 {b}^{2} + 13 \ge 0$
$0 \le g \left(\beta\right) \le 4$

for

$- \sqrt{\frac{1}{2} \left(2 + \sqrt{7}\right)} \le \beta \le \sqrt{\frac{1}{2} \left(2 + \sqrt{7}\right)}$

The conditions for $p \left(\gamma\right)$ are:

$0 \le p \left(\gamma\right) \le 4$

giving

$- 6 \le \gamma \le 18$

Attached a plot showing the variety

$f \left(\alpha\right) + g \left(\beta\right) + p \left(\gamma\right) = 4$ 