How do you solve #2x ^ { 2} - 3x + 8= - x ^ { 2} + 7x#?

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Answer 1 · 2016-12-28T14:38:28

#x=4/3 or 2#

#2x^2-3x+8=-x^2+7x#

or, #2x^2+x^2-3x+8=cancel(-x^2+x^2)+7xrarr#Adding #x^2# to both

sides.

or, #3x^2-3x+8=7x#

or, #3x^2-3x-7x+8=7x-7xrarr#Subtracting #7x# from both sides.

or, #3x^2-10x+8=0rarr#Factor LHS.

or, #3x^2-6x-4x+8=#

or, #3x(x-2)-4(x-2)=0#

or, #(3x-4)(x-2)=0#

Either #3x-4=0#

or, #3x=4#

or, #x=4/3#

OR #x-2=0#

or, #x=2#