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

Jun 26, 2018

$\left(- x + 1\right) \left(x + 3\right) = 0$

$- x + 1 = 0 \mathmr{and} x + 3 = 0$

$x = 1 \mathmr{and} x = - 3$

Jun 26, 2018

$\left(- x + 1\right) \left(x + 3\right) = 0$

Set $- x + 1 = 0 \implies x = 1$
Set $+ x + 3 = 0 \implies x = - 3$

Taken you through the process step by step

#### Explanation:

Given: $- {x}^{2} - 2 x + 3 = 0$

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

Notice that $1 \times 3 = 3 \mathmr{and} - 3 + 1 = - 2$

So in error write: $\left(x - 3\right) \left(x + 1\right) \textcolor{red}{\leftarrow \text{ Does not give } - {x}^{2}}$

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

$\left(- x - 3\right) \left(x + 1\right) = - {x}^{2} \textcolor{red}{- 3 x - x} - 3$

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

(-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 $\left(\textcolor{m a \ge n t a}{+ 1}\right) \times \left(+ 3\right)$

color(green)((-xcolor(magenta)(+1))(x+3)color(white)("dd") =color(white)("dd") -x^2color(white)("d")ubrace(-3x+x)+3 =0
$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddddddddddddd")-x^2color(white)("dd")-2xcolor(white)("dd}} + 3 = 0}$

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So the factorisation is $\left(- x + 1\right) \left(x + 3\right) = 0$

Set $- x + 1 = 0 \implies x = 1$
Set $+ x + 3 = 0 \implies x = - 3$

Jun 26, 2018

x = 1, and x = -3

#### Explanation:

$- {x}^{2} - 2 x + 3 = 0$
Since a + b + c = 0, use shortcut.
The 2 real roots are:
x = 1 , and $x = \frac{c}{a} = \frac{3}{-} 1 = - 3.$

Rule of Shortcut.
If a + b + c = 0, The 2 real roots are: x = 1, and $x = \frac{c}{a}$
If a - b + c = 0, The 2 real roots are: x = -1, and $x = - \frac{c}{a}$

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