# How do you solve 2x²-x+8=0?

##### 3 Answers
Jun 12, 2018

See a solution process below:

#### Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{8}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 1} \pm \sqrt{{\textcolor{b l u e}{- 1}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{8}\right)}}{2 \cdot \textcolor{red}{2}}$

$x = \frac{1 \pm \sqrt{1 - 64}}{4}$

$x = \frac{1 \pm \sqrt{- 63}}{4}$

$x = \frac{1 \pm \sqrt{63 \cdot - 1}}{4}$

$x = \frac{1 \pm \sqrt{63} \sqrt{- 1}}{4}$

The square root of -1 or $\sqrt{- 1}$ is known as an imaginary number and is often referenced as $i$.

Substituting gives the solution set as;

$x = \left\{\frac{1 - \sqrt{63} i}{4} , \frac{1 \pm \sqrt{63} i}{4}\right\}$

Jun 12, 2018

No real roots

#### Explanation:

$y = 2 {x}^{2} - x + 8 = 0$
Use the improved quadratic formula (Socratic, Google Search).
$D = {d}^{2} = {b}^{2} - 4 a c = 1 - 64 = - 63$
Since D < 0, there are no real roots.
There are 2 complex roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a}$, with $d = \pm i \sqrt{63}$
$x = \frac{1}{4} \pm \frac{i \sqrt{63}}{4} = \frac{1 \pm i \sqrt{63}}{4}$

Jun 12, 2018

See below..

#### Explanation:

Let's complete the square:

$2 \left[{x}^{2} - \frac{1}{2} x\right] + 8 :$

$\Rightarrow 2 {\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} + \frac{128}{16}$

$\Rightarrow 2 {\left(x - \frac{1}{4}\right)}^{2} + \frac{127}{16}$

$\Rightarrow 2 {\left(x - \frac{1}{4}\right)}^{2} = - \frac{127}{16}$

$\Rightarrow {\left(x - \frac{1}{4}\right)}^{2} = - \frac{127}{32}$

$x - \frac{1}{4} = \pm \sqrt{- \frac{127}{32}}$

$x = \frac{1}{4} \pm \sqrt{- \frac{127}{32}}$

No real roots as negative square root, although you could use $i$, but I don't know how.