How do you solve using completing the square method #(-2/3)x^2 + (-4/3)x + 1 = 0#?

1 Answer
Shwetank Mauria
Jun 6, 2017

#x=-1+sqrt(5/2)# or #-1-sqrt(5/2)#

Explanation:

We have #(-2/3)x^2+(-4/3)x+1=0#

As #(-2/3)x^2# is not a complete square, let us multiply each term by #-3/2#, so that we get #x^2#, which is a complete square. Then our equation becomes

#(-2/3)xx(-3/2)x^2+(-4/3)xx(-3/2)x-3/2=0#

or #x^2+2x-3/2=0#

or #(x^2+2x+1)-1-3/2=0#

or #(x+1)^2-5/2=0#

or #(x+1)^2-(sqrt(5/2))^2=0#

i.e. #(x+1-sqrt(5/2))(x+1+sqrt(5/2))=0#

Hence, #x=-1+sqrt(5/2)# or #-1-sqrt(5/2)#

Related questions
See all questions in Completing the Square
Impact of this question
1166 views around the world
You can reuse this answer
Creative Commons License