How do you solve #18x ^ { 2} - 42= 120x#?

1 Answer
Jim G.
Oct 12, 2017

#x=-1/3" or "x=7#

Explanation:

#"rearrange terms and equate to zero"#

#rArr18x^2-120x-42=0#

#"take out a "color(blue)"common factor "" of "6#

#rArr6(3x^2-20x-7)=0#

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

#"factors of the product - 21 which sum to - 20 are - 21 and + 1"#

#6(3x^2-21x+x-7)=0larrcolor(blue)" split middle term"#

#6(color(red)(3x)(x-7)color(red)(+1)(x-7))=0larrcolor(blue)" grouping"#

#"factor out "(x-7)#

#6((x-7)(color(red)(3x+1)))=0#

#rArr6(x-7)(3x+1)=0#

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

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

#x-7=0rArrx=7#

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