# How do you solve (x+1)^2-9/4=0?

Feb 6, 2017

$x = - \frac{5}{2} \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = + \frac{1}{2}$

#### Explanation:

${\left(x + 1\right)}^{2} - \frac{9}{4} = 0$

$\rightarrow {\left(x + 1\right)}^{2} = \frac{9}{4}$

$\rightarrow x + 1 = \pm \frac{3}{2}$

$\rightarrow x = - 1 \pm \frac{3}{2}$

$\rightarrow x = - \frac{5}{2} \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = + \frac{1}{2}$

Feb 6, 2017

Being a second-order equation, their are two solutions

$x = \frac{1}{2}$ and $x = - \frac{5}{2}$

#### Explanation:

First, move the right-most term to the right side of the equation

${\left(x + 1\right)}^{2} = \frac{9}{4}$

Now, take the square root of each side. Don't forget that there will be two roots to the right side:

$x + 1 = \pm \frac{3}{2}$

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

$x = - \frac{3}{2} - 1$ or $x = - \frac{5}{2}$

Feb 6, 2017

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

#### Explanation:

While you could multiply out the bracket and then simplify and factorise the quadratic trinomial, an easier method is to realize that

${\left(x + 1\right)}^{2} - \frac{9}{4}$ can be factorised as the difference of squares.

${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$

So, if ${\left(x + 1\right)}^{2} - \frac{9}{4} = 0$

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

Consider each factor being equal to 0.

$x + 1 + \frac{3}{2} = 0 \text{ } \rightarrow x = - \frac{5}{2} = - 2 \frac{1}{2}$

$x + 1 - \frac{3}{2} = 0 \text{ } \rightarrow x = \frac{1}{2}$