# Question #ba6ea

Oct 25, 2017

See solving's below....

#### Explanation:

$2 x + \frac{2}{x} = 3$

First we have to find the value of $x$ before looking for what is asked..

$\frac{2 x}{1} + \frac{2}{x} = \frac{3}{1}$

Multiply through with the LCM which is $x$ in this case

$x \left(\frac{2 x}{1}\right) + \cancel{x} \left(\frac{2}{\cancel{x}}\right) = x \left(\frac{3}{1}\right)$

$x \left(2 x\right) + 2 = x \left(3\right)$

$2 {x}^{2} + 2 = 3 x$

$2 {x}^{2} - 3 x + 2 = 0 \to \text{Quadratic Equation}$

Well in solving a Quadratic Equation you have to find the sum and products of the roots..

In this case the possible roots are $4$ and $1$

In the above equation, you multiply $2$ attached to the square of $x$ with the constant..

Hence $2 \times 2 = 4$

Therefore, we are going to look for a possible value that can multiply each other to get $+ 4$ and can either subtract or add each other to get $- 3$
Hence $4 \mathmr{and} 1$ are the roots..

Given that $\Rightarrow 4 \times 1 = 4 \mathmr{and} - 4 + 1 = - 3$

Now Solving...

$2 {x}^{2} \textcolor{red}{- 4} x \textcolor{red}{+ 1} x + 1 = 0$

$\left(2 {x}^{2} - 4 x\right) \left(- x + 2\right) = 0$

Factorizing..

$2 x \left(x - 2\right) - 1 \left(x - 2\right) = 0$

$\left(x - 2\right) \left(2 x - 1\right) = 0$

$x - 2 = 0$

$x = 2$

$\mathmr{and}$

$2 x - 1 = 0$

$2 x = 1$

$x = \frac{1}{2}$

Hence we should use $x = 2$ since it's positive

Now..

Find, ${x}^{3} + \frac{1}{x} + 2$

Substitute the value of $x$ into the equation..

${x}^{3} + \frac{1}{x} + 2 \to \text{Equation}$

${\textcolor{red}{2}}^{3} + \frac{1}{\textcolor{red}{2}} + 2$

$8 + \frac{1}{2} + 2$

$8 + 2 + \frac{1}{2}$

$10 + \frac{1}{2}$

$10 \frac{1}{2}$