解一元二次方程式 [(x-3)(x+1)+5]/2=(2/3)x^2-(7/6)x,則x=?

2016-01-12 11:37 am

回答 (2)

2016-01-13 12:22 pm
[(x-3)(x+1)+5]/2=(2/3)x^2-(7/6)x
3(x^2 +x-3x-3)+15=4x^2-7x
3x^2 -6x-9+15=4x^2 -7x
x^2 -x-6=0
x=3 or x=-2
2016-01-12 11:43 am
Sol
[(x-3)(x+1)+5]/2=(2/3)x^2-(7/6)x
3*[(x-3)(x+1)+5]=4x^2-7x
3*[(x^2-2x-3)+5]=4x^2-7x
3*(x^2-2x+2)=4x^2-7x
3x^2-6x+6=4x^2-7x
x^2-x-6=0
(x-3)(x+2)=0
x=3 or x=-2


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