# How do you solve (x + 1/3)^2 - 4/9 = 0?

Jun 10, 2016

${\left(x + \frac{1}{3}\right)}^{2} - \frac{4}{9} = 0$
$\implies {\left(x + \frac{1}{3}\right)}^{2} - {\left(\frac{2}{3}\right)}^{2} = 0$

Use the formula named as Difference of Squares", i.e.,
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Here$a = \left(x + \frac{1}{3}\right)$ and $b = \frac{2}{3}$

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

Now, use "zero product property", i.e.,
If $a . b = 0 ,$ then either $a = 0 \mathmr{and} b = 0.$

Here $a = \left(x - \frac{1}{3}\right)$ and $b = \left(x + 1\right)$

$\implies$ either $x - \frac{1}{3} = 0$ or $x + 1 = 0$
$\implies$ either $x = \frac{1}{3}$ or $x = - 1$

Solution Set$= \left\{\frac{1}{3} , - 1\right\}$

Jun 10, 2016

$x = - 1 , \frac{1}{3}$

#### Explanation:

This method does not require factorization into a difference of squares.

Add $\frac{4}{9}$ to both sides of the equation.

${\left(x + \frac{1}{3}\right)}^{2} = \frac{4}{9}$

Take the square root of both sides. Recall that both the positive and negative roots should be taken.

Also note that $\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$.

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

Split this into two equations:

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

color(red)(x=1/3

The negative version:

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

color(red)(x=-1