Question

How do you factor and solve `-x^2-2x+3 = 0`?

Answer 1

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

`-x+1=0 or x+3=0`

`x=1 or x=-3`

Answer 2

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

Set `-x+1=0 => x=1`
Set `+x+3=0=>x=-3`

Taken you through the process step by step

Explanation:

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

First of all I am going to make a deliberate mistake.

Notice that `1xx3=3 and -3+1=-2`

So in error write: `(x-3)(x+1) color(red)(larr" Does not give "-x^2)`

Lets consider just the `x` and ignore the potential error about the constant.

`(-x-3)(x+1) = -x^2color(red)(-3x-x)-3`

Because we have a negative `x` lets use this to give us
the required `-3x+x->-2x`

`(-x+?)(x+?) rarr" we need "color(magenta)(-3x)` so do this:

`color(green)((-x+?)(xcolor(magenta)(+3)) = -x^2-3x+?x+(?xx3))`

We now need `+3` so `?xx3` has to be `(color(magenta)(+1))xx(+3)`

`color(green)((-xcolor(magenta)(+1))(x+3)color(white)("dd") =color(white)("dd") -x^2color(white)("d")ubrace(-3x+x)+3 =0`
`color(green)(color(white)("dddddddddddddddddddd")-x^2color(white)("dd")-2xcolor(white)("dd")+3=0)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the factorisation is `(-x+1)(x+3)=0`

Set `-x+1=0 => x=1`
Set `+x+3=0=>x=-3`

Answer 3

x = 1, and x = -3

Explanation:

`-x^2 - 2x + 3 = 0`
Since a + b + c = 0, use shortcut.
The 2 real roots are:
x = 1 , and `x = c/a = 3/-1 = -3.`

Rule of Shortcut.
If a + b + c = 0, The 2 real roots are: x = 1, and `x = c/a`
If a - b + c = 0, The 2 real roots are: x = -1, and `x = -c/a`

Remember this shortcut. It will save you a lot of time and effort.