How do you solve (x - 2) ^ { 2} + 3= 4?

Apr 28, 2018

$x = 3 \mathmr{and} x = 1$

Explanation:

${x}^{2} - 4 x + 4 + 3 = 4$
then sum all the unknown and the constant which will give you
${x}^{2} - 4 x + 3 = 0$
then factorise the equation and u get
$\left(x - 3\right) \left(x - 1\right) = 0$
therefore $x = 3 \mathmr{and} x = 1$

Apr 28, 2018

$x = 1 \text{ or } x = 3$

Explanation:

$\text{isolate "(x-2)^2" by subtracting 3 from both sides}$

$\Rightarrow {\left(x - 2\right)}^{2} = 4 - 3 = 1$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 2\right)}^{2}} = \pm \sqrt{1} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x - 2 = \pm 1$

$\text{add 2 to both sides}$

$\Rightarrow x = 2 \pm 1$

$\Rightarrow x = 2 - 1 \text{ or } x = 2 + 1$

$\Rightarrow x = 1 \text{ or } x = 3$