# If a number is added to twice its square, the result is 6. How do you find the number?

## List item

Jan 11, 2017

The number could be $1 \frac{1}{2}$ or $- 2$.

#### Explanation:

From the data, taking the number to be $x$, we write:

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

Subtract $6$ from both sides.

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

Factorise.

$2 {x}^{2} + 4 x - 3 x - 6 = 0$

$2 x \left(x + 2\right) - 3 \left(x + 2\right) = 0$

$\left(2 x - 3\right) \left(x + 2\right) = 0$

$2 x - 3 = 0$ or $x + 2 = 0$

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

I get two numbers: $x = - 2 , \frac{6}{4} = \frac{3}{2}$

#### Explanation:

Let's have the unknown number be $x$.

If a number:

$x$

is added to twice it's square:

$x + 2 {x}^{2}$

the result is 6:

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

Now let's find the number:

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(2\right) \left(- 6\right)}}{2 \left(2\right)}$

$x = \frac{- 1 \pm \sqrt{{1}^{2} + 48}}{4}$

$x = \frac{- 1 \pm \sqrt{49}}{4}$

$x = \frac{- 1 \pm 7}{4}$

$x = - 2 , \frac{6}{4} = \frac{3}{2}$