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

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Answer 1 · 2016-06-10T16:35:31

$(x+1/3)^2-4/9=0$
$implies (x+1/3)^2-(2/3)^2=0$

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

Here $a=(x+1/3)$ and $b=2/3$

$implies (x+1/3-2/3)(x+1/3+2/3)=0$
$implies (x+(1-2)/3)(x+(1+2)/3)=0$
$implies (x-1/3)(x+3/3)=0$
$implies (x-1/3)(x+1)=0$

Now, use "zero product property", i.e.,
If $a.b=0,$ then either $a=0 or b=0.$

Here $a=(x-1/3)$ and $b=(x+1)$

$implies$ either $x-1/3=0$ or $x+1=0$
$implies$ either $x=1/3$ or $x=-1$

Solution Set $={1/3,-1}$

Answer 2 · 2016-06-10T16:38:31

$x=-1,1/3$

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

Add $4/9$ to both sides of the equation.

$(x+1/3)^2=4/9$

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

Also note that $sqrt(4/9)=sqrt4/sqrt9=2/3$ .

$x+1/3=+-2/3$

Split this into two equations:

$x+1/3=2/3$

$color(red)(x=1/3$

The negative version:

$x+1/3=-2/3$

$color(red)(x=-1$

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