# How do you complete the square to solve -x^2-2x+3= 0?

Jun 10, 2015

${\left(x + 1\right)}^{2} - 4 = 0$
Solutions: $x = 1 \mathmr{and} x = - 3$

#### Explanation:

If you have to solve the equation you can multiply all by -1:
−x^2−2x+3=0 -> x^2+2x-3=0
It's always more simple to work with positive numbers =)
${x}^{2} + 2 x - 3 = 0$ is similar to a perfect square trinomial that is ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$.
So we can write our equation adding (or subtracting) what is missing to obtain a perfect square:
${x}^{2} + 2 x - 3 = {\left(x + 1\right)}^{2} - 4 = 0$
If you want to solve the equation you have to obtain 4 in the square ${\left(x + 1\right)}^{2} = 4$, so $x + 1 = \pm 2$ and the solutions are:
$x = 1 \mathmr{and} x = - 3$

Jun 11, 2015

$x = 1 , - 3$

#### Explanation:

A perfect square trinomial is in the form ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$.

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

Multiply the equation times $- 1$.

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

The left side of the equation can be made into a perfect square trinomial by completing the square. This will enable the equation to be solved for $x$.

Add $3$ to both sides.

${x}^{2} + 2 x = 3$

Divide the coefficient of the $x$ term by $2$, then square the result. Add it to both sides of the equation.

${\left(\frac{2}{2}\right)}^{2} = 1$

${x}^{2} + 2 x + 1 = 4$

There is now a perfect square trinomial on the left side of the equation.

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$a = x$
$b = 1$

${\left(x + 1\right)}^{2} = 4$

Take the square root of both sides.

$x + 1 = \pm \sqrt{4}$

$x + 1 = \pm 2$

If $x + 1 = 2$, $x = 1$.

If $x + 1 = - 2$, $x = - 3$