3x^2 + 2x - 12 = -7x?

Question

please answer.. math assignment!

wait..
please use solving quadratic equation by completing the square..
the answer should be coming from the difference of 2 squares..
meaning the answer should only be 1..

Answer 1

3x^2+9x-12=0
(3x-3)(x+4)=0
x=1 or -4

Also from 3x^2+9x-12=0 theres a factor of 3 that can be removed

x^2+3x-4=0
(x-1)(x+4)=0
x=1 or -4

Either way both give the same answer

Answer 2

Completing the square

3x² + 2x - 12 = - 7x

3x² + 2x - 12 + 7x = - 7x + 7x

3x² + 9x - 12 = 0

3x² + 9x - 12 + 12 = 0 + 12

3x² + 9x = 12

x² + 3x = 4. . . . Dividing through by 3

x² + 3x +______ = 4 +______

x² + 3x + 9/4 = 4 + 9/4

(x + 3/2)(x + 3/2) =16/4 + 9/4

(x + 3/2)² = 25/4

(√x + 3/2)² = ± √25 / √4

x + 3/2 = ± 5 / 2

x + 3/2 - 3/2 = - 3/2 ± 5/2

x = - 3/2 ± 5/2

- - - - - - - - - - -

Soving for +

x = - 3/2 + 5/2

x = 2/2

x = 1

- - - - - - - - -

Solving for -

x = - 3/2 - 5/2

x = - 8 /2

x = - 4

- - - - - - - - - - - --
Method 2

3x² + 2x - 12 = - 7x

3x² + 2x - 12 + 7x = - 7x + 7x

3x² + 9x - 12 = 0

Find the sum of the middle term. multiply 3 times 12 = 36

Factors of 36 = 2, 3, 4, 6, 9, 12, 18

12 and - 3 satisfy the sum of the middle term

3x² + 9x - 12 = 0

3x² + 12x - 3x - 12 = 0

3x(x + 4) - 3(x + 4)

(3x - 3)(x + 4)

- - - - - -

Roots

3x - 3 = 0

3x - 3 + 3 = 0 + 3

3x = 3

3x / 3 = 3 / 3

x = 3 / 3

x = 1

- - - - - - - -

Roots

x + 4 = 0

x + 4 - 4 = 0 - 4

x = - 4

- - - - - - - - - - - -s-

Answer 3

May be a little longer:

3x²+2x-12 = -7x

1) Get x on one site:

3x²+9x-12 = 0

2) Divide by 3:

x²+3x-4= 0

3) Try to find an x , that fits. 4 must be divided by this x without
of a rest: It might be 1,- 1, 2,- 2,4,-4

Try 1 1²+3*1-4 =0 that is true. So 1 is our x.

4) Now divide x²+3x-4 by (x-1) (Do a normal division) :

x²+3x-4 : (x-1) = x-2
x² - x
-------
2x-4
2x-4
-------
0

5) Now we know:

x²+3x-4= (x-1)*(x-2) and x²+3x-4 = 0

so (x-1)*(x-2) =0

6) When a product =0 each factor is =0 Because of this we
know:

x-1 =0 and x-2 =0

7) So x =1 or x =2

Answer 4

3x²+2x-12 = -7x

3x²+2x+7x-12=0

(3x-3) (x+4)=0

x²+9x-12=0

ax²+bx-c=0

Using a Quadratic Formula:

x=[-b+√b² - 4ac]/2a so substitute the given values,

=[ -9+√ 9²-4(3)(-12)]/2(3)

= [-9 +√81+144]/6

=[-9+√225]/6

=[-9+15]/6

=6/6

x = 1 .................ans (1)

x=[-b+√b² - 4ac]/2a

x=[ -9-√ 9²-4(3)(-12)]/2(3)

=[-9-√225]/6

=[-9-15]/6

=-24/6

x = - 4...............ans (2)

we have two answers, x = 1

or x = - 4

By checking this we have,

3x² + 2x - 12 = -7x substitute the answer which is 1.

3(1) + 2(1) - 12 = -7(1)

3+2 -12 = -7

5 -12 = -7

-7 = - 7

by checking the other answer which is -4
gives,


3x²+9x-12=0 substitute the number -4

3(-4)² + 9(-4) -12 = 0

3(16) -36-12=0

48-36-12=0

12-12=0


answers are correct!

Answer 5

that equation becomes

3x^2 + 9x - 12 = 0
form of a + b + c = 0
roots are x1 = 1
x2 = c/a = -4

Answer 6

ok here given that,
3x^2 + 2x -12 = -7x

=>3x^2 +9x -12 =0

=>3x^2 +12x -3x -12 =0

=>3x (x+4) - 3(x +4) =0

=>(x+4)(3x-3) =0

so x= -4 or x=1

Answer 7

3 x ² + 9 x - 12 = 0
x ² + 3 x - 4 = 0
( x + 4 ).(x - 1 ) = 0
x = - 4, x = 1

Answer 8

Everyone has the correct answer.

Answer 9

what answerer 1 wrote...

Answer 10

didn't your teacher explain how to solve that?