Question

How do you solve `-2x ^ { 2} = 2x - 4`?

Answer 1

`x=1,-2`

Explanation:

rearrange to

`ax^2+bx+c=0`

`-2x^2=2x-4`

`=>2x^2+2x-4=0`

divide by 2

`x^2+x-2=0`

factorise

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

giving

`x+2=0=>x=-2`

`x-1=0=>x=1`

Answer 2

`x=-2 or x=1`

Explanation:

To solve this, you first need to put everything on one side:

`-2x^2=2x-4`
`2x^2+2x-4=0`

From here, we can factor `2` out:

`2(x^2+x-2)=0`

Dividing by two, we get:

`x^2+x-2=0`

This can then be factorised by finding two numbers that multiply to get `-2` and add to get `1`. Looking at the factors of `2`, you can find these numbers are `-1` and `2`.

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

Now, you have two things multiplied together that equal `0`. This means that one of them has to equal `0`. To find the values of `x` for this to happen, we split the two parts:

`x-1=0`
`therefore x=1`

OR

`x+2=0`
`therefore x=-2`

So, the solution to this equation is `x=-2 or x=1`