Question

How do you solve `x^ { 2} - 3x ^ { 2} - x + 3= 0`?

Answer 1

`x=-3/2` and `x=1`

Explanation:

Let's simplify our `x^2` terms to get

`-2x^2-x+3=0`

To make the leading coefficient positive, we can factor out a negative one. We now have

`-color(blue)((2x^2+x-3))=0`

What I have in blue, we can factor by grouping. Here, I will split up the `b` term so we can factor easier. We get

`-(2x^2color(purple)(-2x+3x)-3)=0`

Notice, what I have in purple is the same as `x`, so I did not change the value of this expression.

`-(color(red)(2x^2-2x)+color(lime)(3x-3))=0`

We can factor out a `2x` from the red term, and a `3` from the green term. We now have

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

Both terms have an `x-1` in common, so we can factor that out. We now have

`-color(orchid)((2x+3))color(magenta)((x-1))=0`

Let's be mindful of the negative out front. We can set both factors equal to zero. We get

`-(2x+3)=0`

`-2x-3=0`

`=>-2x=3`

`color(orchid)(=>x=-3/2)`

`-(x-1)=0`

`=>-x+1=0`

`=>-x=-1`

`color(magenta)(=>x=1)`

Our solutions to this quadratic are

`x=-3/2` and `x=1`

Hope this helps!

Answer 2

`x=-3/2 and x =1`

Explanation:

`x^ { 2} - 3x ^ { 2} - x + 3= 0`

`-2x^2-x+3`

`-(x - 1) (2 x + 3)`

`x=-3/2 and x =1`

Answer 3

`x = 3 or x=1 or x=-1`

Explanation:

I suspect there may have been a typing error and question was supposed to have been:

`x^3 -3x^2 -x+3=0`

`(x^3-3x^2) +(-x+3)=0" "larr` make two groups

`x^2(x-3) -1(x-3)=0" "larr` take out common factors

`(x-3)(x^2-1)=0" "larr`factorise

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

Set each factor equal to `0` and solve:

`x-3 = 0" " rarr x= 3`
`x+1=0" "rarr x=-1`
`x-1=0" "rarr x =1`