Solve this?

Question

(solving quadratic equations using factors)
......

6(x^+x) - 2x(x-3) = 0

...
and type out fully, so i can understand please lol


(^=squared)

the answer should be 0, -3

Answer 1

Let x2 mean x squared. I think you mean

6(x2 + x) - 2x(x-3) =

6x2 + 6x - 2x2 + 6x =

4x2 + 12x= 0

So

x2 + 3 x =0

x(x+3)=0

x =0 or -3

Answer 2

x² = x+6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 So x = 3 or - 2 x² + 2x = 0 x(x + 2) = 0 So x = 0 or - 2 60 = x² - 4x x²- 4x - 60 = 0 (x - 10)(x + 6) = 0 So x = 10 or -6 4x = x² - 40 5 x² - 4x - 40 5 = 0 (x - 9)(x + 5) = 0 So x = 9 or -5 -6x = x² + 9 x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 (x + 3)² = 0 So x = -3 -15=x² + 8x x² + 8x +15 = 0 (x + 5)(x + 3) = 0 So x = -5 or -3 -9x = x² + 14 x² + 9x + 14 = 0 (x + 7)(x + 2) = 0 So x = -7 or - 2 wish that enables. =D

Answer 3

6(x^2+x) - 2x(x-3) = 0
6x^2 + 6x - 2x^2 + 6x = 0
4x^2 + 12x = 0
4x(x + 3) = 0 --------- taking out the common factors of 4x.
x = 0 or x+3 = 0
x = 0 or -3

Answer 4

6x^ +6x - 2x^ +6x = 0 multiply it all out.
4x^ + 12x = 0 combine terms
4x(x+3) = 0 factor out common factors
4x = 0 and x+3 = 0
x = 0 x = -3

Answer 5

OK try this:

6(x^2+x) - 2x(x-3) = 0
= 6x^2 + 6x -2x^2 + 6x = 0
= 4x^2 + 12x = 0
Therefore, 4x(x + 3) = 0
x = 0 or x = -3

Hope this helps and if it does vote for it!!

Answer 6

perform indicated operations:

6x^ +6x -2x^+6x =0
4x^ +12x=0
factor out a 4x
4x(x +3)=0
you can solve this by setting each factor equal to zero
one solution is:
4x=0
x=0
another solution is
x+3 = 0
x=-3

Answer 7

>6(x^+x) - 2x(x-3) = 0 ---> distribute
>6x^+6x-2x^+6x=0 ---> combine like terms
>4x^+12x=0 ---> use factoring with common factors(4x is the common factor)
>4x(x+3)=0 ---> equate to zero
>4x=0 ---> x=0
>(x+3)=0 ---> x=negative 3
*answer:x=0 and x= negative 3

Answer 8

6x² + 6x - 2x² + 6x = 0
4x² + 12x = 0
4x.(x + 3) = 0
x = 0 , x = - 3

Answer 9

6(x^2+x) - 2x(x-3) = 0

6x^2+6x-2x^2+6x=0

4x^2+12x=0

4x(x+3)=0

so;

x=0 or x=-3

Answer 10

Hello

Exactly the same method applies to this question as it did to your last question. It is exactly the same type of question.

You should be able to do this now if you understood what I wrote last time