How do you solve #2x ^ { 2} + 6x - 4= x ^ { 2} - x + 4#?

1 Answer
ali ergin
May 6, 2017

#"Either x=1 or x=-8"#

Explanation:

#2x^2+6x-4=x^2-x+4#

#"let us rearrange the equation above."#

#cancel(2x^2)-cancel(x^2)+cancel(6x)+cancel(x)-4-4=0#

#x^2+7x-8=0#

#x^2+7x-8=(x+8)(x-1)#

#(x+8)(x-1)=0#
#"Either (x+8) or (x-1) must be equal to zero."#
#"if " x+8=0" " rArr" "x=-8 #
#"if "x-1=0" "rArr" "x=1#

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