Question

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a) 2 x la adoua + x - 6=0
b)3 x la adoua +4x - 4 =0
d) 2(x-1)la puterea a doua = 3x+2

Answer 1

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Answer 2

      
$a) \\ 2 x^{2} +x-6=0 \\ \\ x_{12} = \frac{-1 \pm \sqrt{1+4*6*2} }{2*2} = \frac{-1 \pm \sqrt{49} }{4} = \frac{-1 \pm 7 }{4} \\ \\ x_1=\frac{-1 + 7 }{4} =\frac{6 }{4}=\frac{3 }{2} \\ x_2=\frac{-1 - 7 }{4} =\frac{-8 }{4}=-2$


$b) \\ 3 x^{2} +4x-4=0 \\ \\ x_{12} = \frac{-4 \pm \sqrt{16+4*4*3} }{6}= \frac{-4 \pm \sqrt{64} }{6}= \frac{-4 \pm 8 }{6}=\frac{-2 \pm 4 }{3} \\ \\ x_1=\frac{-2 + 4 }{3}=\frac{2 }{3} \\ \\ x_2=\frac{-2 - 4 }{3}=\frac{-6 }{3}=-2$


$c) ---- \\ \\ d) \\ 2(x-1)^2 =3x+2 \\ 2( x^{2} -2x+1) =3x+2 \\ 2 x^{2} -4x+2 =3x+2 \\2 x^{2} -4x+2 -3x-2=0 \\ 2 x^{2} -7x=0 }{4} \\ x(2x-7)=0 \\ x_1=0 \\ x_2= \frac{7}{2}$