Explore: Solving to Complete the Square when 𝑎 ≠ 0, 1 Solution

Solving to Complete the Square when a ≠ 0, 1 Solutions

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 $a = 1,$ but is more common when $a \neq 0, 1 .$

Example 5

Solve by completing the square.

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

Implement	Explain (in Steps)
$3 x^{2} + 4 x + 2 = 0$	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
$x^{2} + \frac{4}{3} x + {(\frac{2}{3})}^{2} = - \frac{2}{3} + {(\frac{2}{3})}^{2}$	Simplify and write $+$ after terms on both sides Calculate value that makes LEFT side of equation a perfect square trinomial

Note

$+ = {(\frac{4}{3 \cdot 2})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$

${(x + \frac{2}{3})}^{2} = - \frac{6}{9} + \frac{4}{9} {(x + \frac{2}{3})}^{2} = - \frac{2}{9}$

${\sqrt{(x + \frac{2}{3})}}^{2} = \pm \sqrt{\frac{- 2}{9}}$

$x + \frac{2}{3} = \frac{\pm i \sqrt{2}}{3}$
$x = \frac{- 2 \pm i \sqrt{2}}{3}$

Write left side of the equation as product of a binomial squared
Simplify right side of equation
Solve for x by taking square root of entire equation
Isolate the variable

Example 6

Solve by completing the square.

$3 x^{2} + 1 = 6 x$

$\begin{array}{rcl} \frac{3}{3} x^{2} - \frac{6}{3} x + \frac{1}{3} & = & 0 \\ x^{2} - 2 x + {(1)}^{2} & = & - \frac{1}{3} + {(1)}^{2} \\ + & = & {(\frac{- 2}{2})}^{2} = {(- 1)}^{2} = 1 \begin{matrix} \end{matrix} \\ {(x - 1)}^{2} & = & \frac{2}{3} \\ \sqrt{{(x - 1)}^{2}} & = & \pm \sqrt{\frac{2}{3}} \\ x - 1 & = & \frac{\pm \sqrt{2}}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3}}) \\ x - 1 & = & \frac{\pm \sqrt{6}}{3} \\ x & = & 1 \pm \frac{\sqrt{6}}{3} \end{array}$