Question

(3x+2)(x-1)+(3x+2)(x+1)=0
(2x-1)(x+3)+(2x-1)(x-3)=0
ajutatima va rog am nevoie urgent

Answer 1

$\displaystyle \it a).(3x+2)(x-1)+(3x+2)(x+1)=0 \\ \\ 3x^2-3x+2x-2+3x^2+3x+2x+2=0 \\ \\ 6x^2+4x=0 \\ \\ a=6,~b=4,~c=0 \\ \\ \Delta=b^2-4ac=4^2-4 \cdot 6 \cdot 0=16-0=16\ \textgreater \ 0 \\ \\ x_1 = \frac{-4+ \sqrt{16} }{2 \cdot 6} = \frac{-4+4}{12} =0 \\ \\ x_2= \frac{-4- \sqrt{16} }{2 \cdot 6} = \frac{-4-4}{12} = \frac{-8}{12} =- \frac{2}{3}$

$\displaystyle \it b). (2x-1)(x+3)+(2x-1)(x-3)=0 \\ \\ 2x^2+6x-x-3+2x^2-6x-x+3=0 \\ \\ 4x^2-2x=0 \\ \\ a=4,~b=-2,~c=0 \\ \\ \Delta=b^2-4ac=(-2)^2-4 \cdot 4 \cdot 0=4-0=4\ \textgreater \ 0 \\ \\ x_1= \frac{2+ \sqrt{4} }{2 \cdot 4} = \frac{2+2}{8} = \frac{4}{8} = \frac{1}{2} \\ \\ x_2= \frac{2- \sqrt{4} }{2 \cdot 4} = \frac{2-2}{8} =0$