# How do you solve (2x+1)^2=(x+2)^2?

May 19, 2017

$x = \pm 1$

#### Explanation:

$\text{expand factors on both sides using the FOIL method}$

$4 {x}^{2} + 4 x + 1 = {x}^{2} + 4 x + 4$

$\text{subtract " x^2" from both sides}$

$4 {x}^{2} - {x}^{2} + 4 x + 1 = \cancel{{x}^{2}} \cancel{- {x}^{2}} + 4 x + 4$

$\Rightarrow 3 {x}^{2} + 4 x + 1 = 4 x + 4$

$\text{subtracting 4x from both sides, gives}$

$3 {x}^{2} + 1 = 4$

$\text{subtract 4 from both sides}$

$3 {x}^{2} + 1 - 4 = 4 - 4$

$\Rightarrow 3 {x}^{2} - 3 = 0 \leftarrow \textcolor{b l u e}{\text{ quadratic equation}}$

$3 \left({x}^{2} - 1\right) = 0 \leftarrow \textcolor{b l u e}{\text{ common factor of 3}}$

$3 \left(x - 1\right) \left(x + 1\right) = 0 \leftarrow \textcolor{b l u e}{\text{ difference of aquares}}$

$\text{equate each factor to zero and solve}$

$x - 1 = 0 \Rightarrow x = 1$

$x + 1 = 0 \Rightarrow x = - 1$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the equation and if both sides are equal then they are the solutions.

$x = 1 \to {\left(2 + 1\right)}^{2} = {\left(1 + 2\right)}^{2} \to \text{ True}$

$x = - 1 \to {\left(- 1\right)}^{2} = {\left(1\right)}^{2} \to \text{ True}$

$\Rightarrow x = \pm 1 \text{ are the solutions}$