(x+2) (x+3)+(x-3) (x-2) -2x(x+1) = 0 , solve the following equation.
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Answered by
0
(x+2)(x+3)=x²+3x+2x+6=x²+6x+6
(x-3)(x-2)=x²-2x-3x+6=x²-5x+6
-2x(x+1)=-2x²-2x
after adding all the values , we get -x+12
(x-3)(x-2)=x²-2x-3x+6=x²-5x+6
-2x(x+1)=-2x²-2x
after adding all the values , we get -x+12
Answered by
3
(X+2) (x+3)+(x-3) (x-2)-2x(x+1)=0
(X^+3x+2x+6)+(x^-2x-3x+6)-2x^-2x=0
X^+3x+2x+6+x^-2x-3x+6-2x^-2x=0
X^+x^-2x^+3x+2x-2x-3x-2x+6+6-2=0
2X^-2x^+5x-5x-2x+12=0
-2x+12=0
-2x=-12
X=12/2
Therefore X=6 ans (^ this singh is square )
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