How do you solve #2x^2 -: 3x = 14# ?

1 Answer
Aug 20, 2016

Answer:

See explanation...

Explanation:

I think the question should have had subtraction rather than division, but I will answer both possibilities:

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How to solve #bb((2x^2)/(3x)=14)#

Multiply both sides by #3x# to get:

#2x^2 = 42x#

Divide both sides by #2x# to find:

#x = 21#

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How to solve #bb(2x^2-3x=14)#

Subtract #14# from both sides to get:

#2x^2-3x-14 = 0#

Use an AC method to factorise:

Look for a pair of factors of #AC=2*14=28# which differ by #B=3#.

The pair #7, 4# works.

Use this to split the middle term then factor by grouping:

#0 = 2x^2-3x-14#

#=2x^2-7x+4x-14#

#=(2x^2-7x)+(4x-14)#

#=x(2x-7)+2(2x-7)#

#=(x+2)(2x-7)#

Hence #x=-2# or #x=7/2#