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

2 Answers
Jun 10, 2016

#(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}#

Jun 10, 2016

Answer:

#x=-1,1/3#

Explanation:

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#