How do you solve the quadratic equation by completing the square: 3x^2+9x-1=0?

Jul 16, 2015

$x = - \frac{3}{2} \pm \frac{\sqrt{31}}{2 \sqrt{3}}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(by completing the square)

Explanation:

Given $3 {x}^{2} + 9 x - 1 = 0$

Move the constant to the right side and extract the common factor of $3$ from the left side:
$\textcolor{w h i t e}{\text{XXXX}}$$3 \left({x}^{2} + 3 x\right) = 1$

If ${x}^{2} + 3 x$ are the first two terms of the expansion of a squared binomial:
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + a\right)}^{2} = \left({x}^{2} + 2 a x + {a}^{2}\right)$

then $a = \frac{3}{2}$ and ${a}^{2} = {\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$
$\textcolor{w h i t e}{\text{XXXX}}$which is the amount we will need to add (inside the parentheses) to "complete the square)

$\textcolor{w h i t e}{\text{XXXX}}$Note that adding $\frac{9}{4}$ inside the parentheses is the same as adding $3 \cdot \left(\frac{9}{4}\right)$ and this amount will need to be added to the right side as well to keep the equation balanced.

$\textcolor{w h i t e}{\text{XXXX}}$$3 \left({x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2}\right) = 1 + \frac{27}{4}$

Rewriting as a squared binomial (and simplifying the right side)
$\textcolor{w h i t e}{\text{XXXX}}$$3 {\left(x + \frac{3}{2}\right)}^{2} = \frac{31}{4}$

Dividing both sides by $3$
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + \frac{3}{2}\right)}^{2} = \frac{31}{12}$

Taking the square root of both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$x + \frac{3}{2} = \pm \sqrt{\frac{31}{12}}$

Subtracting $3$ from both sides
$\textcolor{w h i t e}{\text{XXXX}}$$x = - \frac{3}{2} \pm \frac{\sqrt{31}}{2 \sqrt{3}}$