# How do you solve using the completing the square method x^2 + 2x - 15 = 0?

Apr 4, 2016

The solutions are:
color(green)(x =3 or  color(green)(x = -5

#### Explanation:

${x}^{2} + 2 x - 15 = 0$

${x}^{2} + 2 x = 15$

To write the Left Hand Side as a Perfect Square, we add 1 to both sides:

${x}^{2} + 2 x + 1 = 15 + 1$

${x}^{2} + 2 \cdot x \cdot 1 + {1}^{2} = 16$

Using the Identity color(blue)((a+b)^2 = a^2 + 2ab + b^2, we get

${\left(x + 1\right)}^{2} = 16$

$x + 1 = \sqrt{16}$ or $x + 1 = - \sqrt{16}$

color(green)(x = 4 - 1 =3 or  color(green)(x = 4 - 1 = -5

Apr 4, 2016

$\textcolor{b l u e}{\implies + x = - 1 \pm 4}$

$\textcolor{b r o w n}{x = + 3 \text{ or } - 5}$

#### Explanation:

The equation in standard form is:
$\text{ } \textcolor{g r e e n}{y = a {x}^{2} + b x + c}$

The equation in vertex form is:
$\text{ } \textcolor{b l u e}{y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]}$

color(brown)("The "-[(b/2)^2]" corrects the error produced by" (b/(2a))^2

They are different versions of the same thing!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Solving your question}}$

$a = 1 \text{ ; "b=2" ; } c = - 15$ giving

$y = {\left(x + 1\right)}^{2} - 15 - 1$

$\textcolor{b l u e}{y = {\left(x + 1\right)}^{2} - 16}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set the vertex form equation as equal to zero

$\implies 0 = {\left(x + 1\right)}^{2} - 16$

$\implies {\left(x + 1\right)}^{2} = + 16$

Taking the square root of both sides

$\implies \pm \left(x + 1\right) = \pm \sqrt{16}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider the case } - \left(x + 1\right)}$

$\implies - x - 1 = \pm 4$

$\implies - x = + 1 \pm 4$

Multiply both sides by (-1)

$\textcolor{b l u e}{\implies + x = - 1 \pm 4}$

$\textcolor{b r o w n}{x = + 3 \text{ or } - 5}$

'....................................................................

$\textcolor{b l u e}{\text{Consider the case } + \left(x + 1\right)}$

$\implies x = - 1 \pm 4$

$\textcolor{b r o w n}{\text{Same as for } - \left(x + 1\right)}$
'...................................................