Solving to Complete the Square when a ≠ 0, 1

• When solving problems by completing the square, the result may included    complex solutions   .
  • When complex solutions occur, remember to first    simplify –1 out    of any radical, and check that the denominator is    rationalized   .
  • Complex solutions can occur when, but is more common when $\underline{ \mathit{a}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{ }\mathbf{ }},$ but is more common when $\underline{ \mathit{a}\mathbf{\ne }\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{ }\mathbf{ }}.$

Example 5

Solve by completing the square.

$3x^2+4x+2=0$

Implement	Explain (in Steps)
$3x^2+4x+2=0$	Write equation in standard form
$\frac{3}{3}{x}^2+\frac{4}{3}x+\frac{2}{3}=0$	Divide all terms by leading coefficient, a (If $a=1$ this step can be skipped)
$x^2+\frac{4}{3}x=–\frac{2}{3}$	Add $–\frac{c}{a}$ to both sides of equation
$x^2+\frac{4}{3}x+\boxed{ {\left(\frac{2}{3}\right)}^2 }=–\frac{2}{3}+\boxed{ {\left(\frac{2}{3}\right)}^2 }$	Simplify and write $+\boxed{ }$ after terms on both sides Calculate value that makes LEFT side of equation a perfect square trinomial

Note

$+\boxed{ }={\left(\frac{4}{3·2}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$

$$\begin{gathered} {\left(x+\frac{2}{3}\right)}^2=–\frac{6}{9}+\frac{4}{9} \\ {\left(x+\frac{2}{3}\right)}^2=–\frac{2}{9} \end{gathered}$$	Write left side of the equation as product of a binomial squared: ${\left(x–\frac{b}{2}\right)}^2$ Simplify right side of equation
$\sqrt{{\left(x+\frac{2}{3}\right)}^2}=\pm \sqrt{\frac{–2}{9}}$	Solve for x by taking square root of entire equation
$$\begin{gathered} x+\frac{2}{3}=\frac{\pm i\sqrt{2}}{3} \\ x=\frac{–2\pm i\sqrt{2}}{3} \end{gathered}$$	Isolate the variable

Example 6

Solve by completing the square.

$3x^2+1=6x$

$\begin{array}{rcl}\frac{3}{3}{x}^2–\frac{6}{3}x+\frac{1}{3}& =& 0\\ x^2–2x+ \boxed{ {\left(1\right)}^2 } & =& –\frac{1}{3}+\boxed{ {\left(1\right)}^2 }\\ +\boxed{ }& =& {\left(\frac{–2}{2}\right)}^2={\left(–1\right)}^2=1\begin{array}{}\end{array}\\ {\left(x–1\right)}^2& =& \frac{2}{3}\\ \sqrt{{\left(x–1\right)}^2}& =& \pm \sqrt{\frac{2}{3}}\\ x–1& =& \frac{\pm \sqrt{2}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)\\ x–1& =& \frac{\pm \sqrt{6}}{3}\\ x& =& 1\pm \frac{\sqrt{6}}{3}\\ & & \end{array}$