# How do you solve (x + 3) ² – (2x – 1)1 = 0?

Jan 27, 2018

$\therefore x = - 2$ (plus or minus) $24 i$

#### Explanation:

Rewrite Question:

$\left(x + 3\right) \left(x + 3\right) - \left(2 x - 1\right) 1 = 0$

Distribute:

$\left({x}^{2} + 3 x + 3 x + 9\right) - \left(2 x - 1\right)$

Collect Like Terms:

${x}^{2} + 6 x + 9 - 2 x + 1 = 0$

${x}^{2} + 4 x + 10 = 0$

There are no REAL SOLUTIONS .

You can not square root a NEGATIVE NUMBER.

Therefore:

$\therefore x = - 2$ (plus or minus) $24 i$

Jan 27, 2018

There are two imaginary solutions:

$x = - 2 + i \sqrt{6} , - 2 - i \sqrt{6}$

#### Explanation:

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

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

${x}^{2} + 6 x + 9 - 2 x + 1 = 0$

${x}^{2} + 4 x + 10 = 0$

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

$\implies x = \frac{- 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(10\right)}}{2 \left(1\right)}$

$= \frac{- 4 \pm \sqrt{16 - 40}}{2}$

$= \frac{- 4 \pm \sqrt{- 24}}{2}$

Since the number under the square root is negative, the solutions are imaginary. You could write "no real solution" for the answer, or if you want the imaginary ones:

$\frac{- 4 \pm \sqrt{- 24}}{2}$

$= \frac{- 4 \pm \sqrt{- 4 \cdot 6}}{2}$

$= \frac{- 4 \pm 2 i \sqrt{6}}{2}$

$= - 2 \pm i \sqrt{6}$

Here are the final imaginary solutions:

$x = - 2 + i \sqrt{6} , - 2 - i \sqrt{6}$