# How do you solve 6x² + x = 2?

May 1, 2016

The solutions are:

color(blue)(x= 1/2

color(blue)(x= -2/3

#### Explanation:

$6 {x}^{2} + x = 2$

$6 {x}^{2} + x - 2 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 6 , b = 1 , c = - 2$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(1\right)}^{2} - \left(4 \cdot 6 \cdot \left(- 2\right)\right)$

$= 1 + 48 = 49$

The solutions are found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 1\right) \pm \sqrt{49}}{2 \cdot 6} = \frac{- 1 \pm 7}{12}$

$x = \frac{- 1 + 7}{12} = \frac{6}{12} = \frac{1}{2}$

$x = \frac{- 1 - 7}{12} = - \frac{8}{12} = - \frac{2}{3}$

May 1, 2016

The method of finding the solution will depend on the grade in which it is being dealt with. An alternative method is to factorise.
$x = \frac{1}{2}$ or $x = - \frac{2}{3}$

#### Explanation:

To solve a quadratic $\left({x}^{2}\right)$ equation, always make it equal to 0.

In simple algebra there are then three methods available to find the solution: factorise, complete the square or the quadratic formula.

$6 {x}^{2} + x - 2 = 0$ factorises into $\left(2 x - 1\right) \left(3 x + 2\right) = 0$

The fact that a product of two factors results in 0 means that at least one of the factors must be equal to 0.

If $2 x - 1 = 0 ,$ then $2 x = 1 \mathmr{and} x = \frac{1}{2}$

If $3 x + 2 = 0 ,$ then $3 x = - 2 \mathmr{and} x = - \frac{2}{3}$