Checkpoint: Solving to Complete the Square when 𝑎 = 1 Solution

Solving to Complete the Square when a = 1 Solution

It is very important that your student completes this teach-back, demonstrating they can verbalize the steps for completing the square.

$x^{2} + 3 x + 2 = - 2 x$	(Original problem for reference.)
Worked solution for Example 4	Student should explain:
$x^{2} + 5 x + 2 = 0$	Write the equation in standard form.
$x^{2} + 5 x = - 2$	Divide all terms by leading coefficient, a (If $a = 1$ this step can be skipped) $a = 1$ Continue to step 3.
$x^{2} + 5 x + {(\frac{5}{2})}^{2} = - 2 + {(\frac{5}{2})}^{2}$	Add $- \frac{c}{a}$ to both sides of equation
${(x + \frac{5}{2})}^{2} = - 2 + \frac{25}{4} {(x + \frac{5}{2})}^{2} = - \frac{8}{4} + \frac{25}{4}$	Simplify and write $+$ after terms on both sides Calculate value that makes LEFT side of equation a perfect square trinomial

Note

+ = {(\frac{5}{2})}^{2} = \frac{25}{4}

This step can be completed using mental math. Therefore, students may or may not include this in their explanation.

If work is shown, it is often done off to the side so that the rest of the problem flows together.

Remember to add the value

{(\frac{b}{2})}^{2}

to the blanks on both sides of the equation.

$\sqrt{{(x + \frac{5}{2})}^{2}} = \pm \sqrt{\frac{17}{4}}$	Write the left side of the equation as a product of a binomial squared: ${(x - \frac{b}{2})}^{2}$ Simplify the right side of the equation.
$x + \frac{5}{2} = \pm \frac{\sqrt{17}}{2}$	Solve for x by taking the square root of the entire equation.
$x = - \frac{5}{2} \pm \frac{\sqrt{17}}{2}$	Isolate the variable.