Solve the equation.
(2r + 1 )( r -2 ) = - (3r -6 )
Solve the equation.
(2r + 1 )( r -2 ) = - (3r -6 )
I like answers above, but want to give another approach, this is algebra any way.
(2r + 1)(r - 2) = - (3r - 6)
(2r + 1)(r - 2) + (3r - 6) = 0
(2r + 1)(r - 2) + 3(r - 2) = 0 , factor out (r - 2)
(r - 2)(2r + 1 + 3) = 0
(r - 2)(2r + 4) = 0 , factor out 2
2(r - 2)(r + 2) = 0
Product equal zero if at least one factor equal zero.
r_{1,2} = ±2
(2r + 1)(r - 2) = -(3r - 6)
2r^{2 }- 4r + r - 2 = -3r + 6
2r^{2 }- 3r - 2 = -3r + 6
2r^{2 }- 3r + 3r - 2 - 6 = 0
2r^{2 }- 8 = 0
(2r^{2} - 8) / 2 = 0 / 2
r^{2 }- 4 = 0
r^{2 }= 4
r = ± 2
Expand and simplify,
2r^{2} - 3r - 2 = -3r + 6
r^{2} = 4
r = +/- 2 <==Answer
(2r + 1 )( r -2 ) = - (3r -6 )
2r^{2}+ r-4r-2 = -3r+6
2r^{2}-3r-2=-3r+6
2r^{2}=8
r^{2}=4
Thus r= +2 or -2
2r^2 + r - 4r -2 + 3r -6 = 0 = 2r^2 - 8 = (2r + 4)(r - 2)
r = 2 or -2