How do you solve #12x^2-14x+4=0#?

1 Answer
Jim G.
May 8, 2018

#x=2/3" or "x=1/2#

Explanation:

#"simplify by dividing through by 2"#

#rArr6x^2-7x+2=0#

#"using the a-c method for factorising"#

#"the factors of the product "6xx2=12#

#"which sum to - 7 are - 3 and - 4"#

#"split the middle term using these factors"#

#6x^2-3x-4x+2=0larrcolor(blue)"factor by grouping"#

#color(red)(3x)(2x-1)color(red)(-2)(2x-1)=0#

#"take out the "color(blue)"common factor "(2x-1)#

#rArr(2x-1)(color(red)(3x-2))=0#

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

#3x-2=0rArrx=2/3#

#2x-1=0rArrx=1/2#

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