Question

PLease help me with this question ill give you points

Answer 1

Factorise :

`\qquad \sf \: { \dashrightarrow {6x}^(2) + 4x= 2 - 7x}`

`\qquad \sf \: { \dashrightarrow {6x}^(2) + 4x + 7x - 2 =0}`

`\qquad \sf \: { \dashrightarrow {6x}^(2) + 11x - 2 =0}`

`\qquad \sf \: { \dashrightarrow {6x}^(2) + 12x - x - 2 =0}`

`\qquad \sf \: { \dashrightarrow {6x}^{} (x+ 2) -1( x + 2 )=0}`

`\qquad \sf \: { \dashrightarrow ({6x}^{} - 1) (x+ 2) =0}`

`\qquad \sf \: { \therefore 6x - 1=0}`

`\qquad \sf \: \dashrightarrow6x =1`

`\qquad \sf \: { \dashrightarrow x= (1)/(6) }`

`\qquad \sf \: { \therefore x + 2=0}`

`\qquad \sf \: { \dashrightarrow x = - 2} \\`

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

Answer 2

`x = -2, x = (1)/(6)`

completing square method:

• 
  `\rf 6x^2 + 4x = 2 -7x`

• 
  `\rf 6x^2 + 4x + 7x -2 = 0`

• 
  `\rf 6x^2 + 11x -2 = 0`

• 
  `\rf 6x^2 +12x - x - 2 = 0`

• 
  `\rf 6x ( x + 2) -1(x + 2) = 0`

• 
  `\rm (x+2)(6x-1) = 0`

• 
  `x = -2, x = (1)/(6)`