# How do you solve 3x^2+2=4x ?

Jul 29, 2016

$\implies x = \frac{2}{3} + \frac{\sqrt{3}}{3} i \text{ }$ and $\text{ } x = \frac{2}{3} - \frac{\sqrt{3}}{3} i$

$x \notin \mathbb{R} \text{ but } x \in \mathbb{C}$

#### Explanation:

If able to do so the quickest way is to factorise.

Subtract $4 x$ from both sides giving:

$3 {x}^{2} - 4 x + 2 = 0$

Try 1
$\left(3 x + 2\right) \left(x + 1\right) = 3 {x}^{2} + 3 x + 2 x + 2 \textcolor{red}{\leftarrow \text{ Fail}}$

Try2
$\left(3 x + 1\right) \left(x + 2\right) = 3 {x}^{2} + 6 x + 1 x + 2 \textcolor{red}{\leftarrow \text{ Fail}}$

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Looks as though we need to use another approach

$\textcolor{b l u e}{\text{Completing the square}}$

For detailed method of approach look at
https://socratic.org/s/awA8fpNk

Given:$\text{ } 3 {x}^{2} - 4 x + 2 = 0$

Write as: $3 \left({x}^{2} - \frac{4}{3} x\right) + 2 + k = 0$

Giving:$\text{ } 3 {\left(x - \frac{4}{6}\right)}^{2} + 2 - \frac{4}{3} = 0$

$\implies 3 {\left(x - \frac{2}{3}\right)}^{2} + \frac{2}{3} = 0$
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${\left(x - \frac{2}{3}\right)}^{2} = - \frac{2}{3} \times \frac{1}{3} \equiv - \frac{1}{3}$

$x - \frac{2}{3} = \sqrt{- \frac{1}{3}}$

Square root of a negative number means that the curve does not cross the x-axis so there is no solution in the set of numbers called 'Real'. Written as $x \notin \mathbb{R}$

However there is at least 1 solution in the set of number called 'Complex'. Written as $x \in \mathbb{C}$
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So we now have:

$x = \frac{2}{3} \pm \frac{1}{\sqrt{3}} \sqrt{- 1}$

$\implies x = \frac{2}{3} + \frac{\sqrt{3}}{3} i \text{ }$ and $\text{ } x = \frac{2}{3} - \frac{\sqrt{3}}{3} i$