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How do you solve $3 x^{3} - 2 [x - (3 - 2 x)] = 3 {(x - 2)}^{2}$ $3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2$ ?

1 Answer

Jun 16, 2016

Real solution: $x = 1$ $x=1$

Complex solutions: $x \in {1, \sqrt{2} i, - \sqrt{2} i}$ $x in {1, sqrt(2)i, -sqrt(2)i}$

Explanation:

L.S. $= 3 x^{3} - 2 [x - (3 - 2 x)]$ $=3x^3-2[x-(3-2x)]$

$XXX = 3 x^{3} - 2 [- 3 + 3 x]$ $color(white)("XXX")=3x^3-2[-3+3x]$

$XXX = 3 x^{3} - 6 x + 6$ $color(white)("XXX")=3x^3-6x+6$

R.S. $= 3 {(x - 2)}^{2}$ $=3(x-2)^2$

$XXX = 3 (x^{2} - 4 x + 4)$ $color(white)("XXX")=3(x^2-4x+4)$

$XXX = 3 x^{2} - 12 x + 12$ $color(white)("XXX")=3x^2-12x+12$

Since L.S. $=$ $=$ R.S.
$3 x^{3} + 6 x + 6 = 3 x^{2} - 12 x + 12$ $3x^3+6x+6=3x^2-12x+12$

$\to x^{3} + 2 x + 2 = x^{2} - 4 x + 4$ $rarr x^3+2x+2=x^2-4x+4$

$\to 1 x^{3} - x^{2} + 2 x - 2 = 0$ $rarr color(green)(1)x^3-x^2+2x-color(blue)(2)=0$

In the hope of finding rational roots we try the factors of $\frac{2}{1}$ $color(blue)(2)/color(green)(1)$
enter image source here
and we find $x = 1$ $color(red)(x=1)$ is a root
$\to (x - 1)$ $rarr (x-1)$ is a factor.

$(x^{3} - x^{2} + 2 x - 2) \div (x - 1) = x^{2} + 2$ $(x^3-x^2+2x-2) div (x-1) = x^2+2$

$(x^{2} + 2)$ $(x^2+2)$ has no Real roots (i.e. no Real solutions for $x$ $x$ )

However, if we allow Complex solutions:
$XXX x = \pm \sqrt{2} i$ $color(white)("XXX")x=+-sqrt(2)i$

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