Respuesta :

fieryanswererft

Answer:

[tex]x = -2, x = \frac{1}{6}[/tex]

completing square method:

  • [tex]\rf 6x^2 + 4x = 2 -7x[/tex]

  • [tex]\rf 6x^2 + 4x + 7x -2 = 0[/tex]

  • [tex]\rf 6x^2 + 11x -2 = 0[/tex]

  • [tex]\rf 6x^2 +12x - x - 2 = 0[/tex]

  • [tex]\rf 6x ( x + 2) -1(x + 2) = 0[/tex]

  • [tex]\rm (x+2)(6x-1) = 0[/tex]

  • [tex]x = -2, x = \frac{1}{6}[/tex]

Factorise :

[tex]\qquad \sf \: { \dashrightarrow {6x}^{2} + 4x= 2 - 7x}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {6x}^{2} + 4x + 7x - 2 =0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {6x}^{2} + 11x - 2 =0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {6x}^{2} + 12x - x - 2 =0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {6x}^{} (x+ 2) -1( x + 2 )=0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow ({6x}^{} - 1) (x+ 2) =0}[/tex]

[tex]\qquad \sf \: { \therefore 6x - 1=0} [/tex]

[tex] \qquad \sf \: \dashrightarrow6x =1 [/tex]

[tex]\qquad \sf \: { \dashrightarrow x= \dfrac{1}{6} }[/tex]

[tex]\qquad \sf \: { \therefore x + 2=0} [/tex]

[tex]\qquad \sf \: { \dashrightarrow x = - 2} \\ [/tex]

  • Therefore, whether the value of x = 1/6, x = –2

Otras preguntas