How do you solve #6x^2+5x-4=0#?

1 Answer
Jim G.
Apr 18, 2018

#x=-4/3" or "x=1/2#

Explanation:

#"using the a-c method of factoring"#

#"the factors of the product "6xx-4=-24"#
#"which sum to + 5 are - 3 and + 8"#

#"split the middle term using these factors"#

#6x^2-3x+8x-4=0#

#rArrcolor(red)(3x)(2x-1)color(red)(+4)(2x-1)=0#

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

#rArr(2x-1)(color(red)(3x+4))=0#

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

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

#3x+4=0rArrx=-4/3#

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