How to solve for x if #x^2+(2/3)x-1/3=0# ?

1 Answer
Jim G.
Apr 10, 2018

#x=-1" or "x=1/3#

Explanation:

#"take out a "color(blue)"common factor "1/3#

#rArr1/3(3x^2+2x-1)=0#

#"factor the quadratic using the a-c method"#

#"the factors of the product "3xx-1=-3" which sum"#
#"to + 2 are + 3 and - 1"#

#"split the middle term using these factors"#

#3x^2+3x-x-1=0larrcolor(blue)"factor in groups"#

#color(red)(3x)(x+1)color(red)(-1)(x+1)=0#

#"take out the common factor "(x+1)#

#rArr(x+1)(color(red)(3x-1))=0#

#"equate each factor to zero and solve for x"#

#x+1=0rArrx=-1#

#3x-1=0rArrx=1/3#

Related questions
Impact of this question
1223 views around the world
You can reuse this answer
Creative Commons License