# Math Help - quadratic equation problem 2

1. ## quadratic equation problem 2

6x^2 - x - 2=0
6x^2 -4x +3x -2=0
2x(3x-2)1(3x-2)=0
(2x+1)=0 2x=-1 ,x=-1/2

3x-2=0,x=-2/3

I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1

2. If x = 1, then 6x^2 - x - 2 = 6(1)^2 - (1) - 2 = 6 - 1 - 3 = 3, not zero.

Either the book has a typo, or maybe you looked at the solution to the wrong exercise...?

3. Originally Posted by wolfhound
6x^2 - x - 2=0
6x^2 -4x +3x -2=0
2x(3x-2)1(3x-2)=0
(2x+1)=0 2x=-1 ,x=-1/2

3x-2=0,x=-2/3

I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1
(2x+1)(3x-2)=0

x=-1/2 , x=2/3

4. Originally Posted by stapel
If x = 1, then 6x^2 - x - 2 = 6(1)^2 - (1) - 2 = 6 - 1 - 3 = 3, not zero.

Either the book has a typo, or maybe you looked at the solution to the wrong exercise...?
Sorry I mean x=-1.33 and x=1 as the final answers
I didnt mean x=1 as the initial value!?

(2x+1)(3x-2)=0

x=-1/2 , x=2/3

I know its Wrong!
That why I have asked for help here????

6. Originally Posted by wolfhound
I know its Wrong!
That why I have asked for help here????
i mean the answer of the book is wrong , not yours but of course yours is wrong too .

i mean the answer of the book is wrong , not yours but of course yours is wrong too .
If mine is wrong then what is the correct way of doing it?

8. Originally Posted by wolfhound
If mine is wrong then what is the correct way of doing it?
I just posted the correct answers . Well , you are almost correct but you added a negative sign to one of the roots . Refer back to my post .

I just posted the correct answers . Well , you are almost correct but you added a negative sign to one of the roots . Refer back to my post .
could you tell me how change the fractions into decimal numbers please?

10. Originally Posted by wolfhound
6x^2 - x - 2=0
6x^2 -4x +3x -2=0
2x(3x-2)+(3x-2)=0 <<<<<< here you made a tiny but very effective typo
(2x+1)=0 2x=-1 ,x=-1/2

3x-2=0,x=-2/3

I cant understand where I went wrong on this the answer is supposed to be x=-1.33 and x=1
$6x^2 -4x +3x -2=0~\implies~2x(3x-2)+(3x-2)=0~\implies~(3x-2)(2x+1)=0$ and so on ...

11. Originally Posted by earboth
$6x^2 -4x +3x -2=0~\implies~2x(3x-2)+(3x-2)=0~\implies~(3x-2)(2x+1)=0$ and so on ...
thanks
could you tell me if x=1/2 or x=2/3 is the correct answer??
and how could I change these out of fractions please?