# How do you solve 3x^5 = 6x^4 - 72x^3 =?

Jul 22, 2016

$\left(1\right) :$ The Soln., in RR, x=0; & in $\mathbb{C} , x = 0 , x = 1 \pm i \sqrt{23}$.

$\left(2\right) :$ The Reqd. Value, (i.e., "=?" of the problem) in RR, =0; and, in $\mathbb{C} , 0 , \mathmr{and} , - 288 \left(- 11 \pm i \sqrt{23}\right)$.

#### Explanation:

Given that, $3 {x}^{5} = 6 {x}^{4} - 72 {x}^{3}$

$\Rightarrow 3 {x}^{5} - 6 {x}^{4} + 72 {x}^{3} = 0$

$\Rightarrow 3 {x}^{3} \left({x}^{2} - 2 x + 24\right) = 0$

$\Rightarrow {x}^{3} = 0 , \mathmr{and} , {x}^{2} - 2 x + 24 = 0$

${x}^{3} = 0 \Rightarrow x = 0. \ldots \ldots \ldots \left(1\right)$

${x}^{2} - 2 x + 24 = 0. \ldots \ldots \ldots \ldots \ldots \left(2\right)$

$\Rightarrow {x}^{2} - 2 x = - 24$

$\Rightarrow {x}^{2} - 2 x + 1 = - 24 + 1$

$\Rightarrow {\left(x - 1\right)}^{2} = - 23$, which is not possible in $\mathbb{R}$

However, in $\mathbb{C} , {\left(x - 1\right)}^{2} = - 23 \Rightarrow x - 1 = \pm i \sqrt{23}$

$\Rightarrow x = 1 \pm i \sqrt{23}$

Hence, in $\mathbb{R}$, the reqd. value (i.e., "=?" of the problem)$= 0$

While, in $\mathbb{C}$, the reqd. value$= 3 {x}^{5} = 0 , \mathmr{if} x = 0$, or,

if, $x = 1 \pm i \sqrt{23}$, the reqd. value (i.e., "=?" of the problem)

$6 {x}^{4} - 72 {x}^{3} = 6 {x}^{2} \left({x}^{2} - 12 x\right) = 6 \left(2 x - 24\right) \left(- 24\right) \ldots \ldots \ldots \left[b y \left(2\right)\right]$

$= - 24 \cdot 6 \cdot 2 \left(x - 12\right) = - 288 \left(1 \pm i \sqrt{23} - 12\right) = - 288 \left(- 11 \pm i \sqrt{23}\right)$