Solve quadratic equations by taking square roots and by completing the square, recognizing when each method applies (TN A1.A.REI.B.4).

What this topic is asking

Standard A1.A.REI.B.4 also covers two more solving methods. The square-root property solves quadratics with no linear term or already in squared form. Completing the square turns any quadratic into squared form, so it works universally and also converts standard form to vertex form. Both methods rely on the plus-or-minus step when you undo a square.

The square-root property

When a quadratic has the form (something squared) equal to a number, isolate the square and take the root of both sides, keeping both signs.

Completing the square

Completing the square rewrites $x^2 + bx + c = 0$ so the variable terms form a perfect square. The pivotal step is adding $\left(\frac{b}{2}\right)^2$ to both sides, which makes the left side factor as $\left(x + \frac{b}{2}\right)^2$.

The routine:

1. Move the constant to the right side.
2. Add $\left(\frac{b}{2}\right)^2$ to both sides.
3. Factor the left side as a binomial squared.
4. Take square roots of both sides, with $\pm$.
5. Solve for $x$.

If the leading coefficient is not $1$, divide every term by it first, because the perfect-square step assumes a leading coefficient of $1$.

How TNReady examines this topic

• Numeric response. Solve a squared equation and enter both solutions.
• Equation response. Complete the square and give answers in simplest radical form.
• Multiple choice. Choose the value added to complete the square, or the vertex form produced.

A clarifying idea is that completing the square is also how you find a parabola's vertex: rewriting $y = x^2 + 6x + 5$ as $y = (x + 3)^2 - 4$ shows the vertex at $(-3, -4)$. The same algebra that solves the equation reveals the graph's turning point.

Why completing the square always works

Factoring fails when no integer factors exist, but completing the square never does, because every quadratic can be forced into the form $(x + p)^2 = q$. That is also why this method is used to derive the quadratic formula: completing the square on the general equation $ax^2 + bx + c = 0$ produces $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, the formula on the reference sheet. So the formula is just the result of completing the square once, in general, which is why both methods give identical answers. On the test, reach for completing the square when an item explicitly asks for it, when you need vertex form, or when the linear coefficient is even (so $\frac{b}{2}$ is a whole number and the arithmetic stays clean); otherwise the formula is often faster.

A leading coefficient other than 1

When $a \ne 1$, divide every term by $a$ before completing the square, so the perfect-square step works cleanly.

When square roots are the fastest method

The square-root property is the quickest route whenever a quadratic has no linear term, that is when $b = 0$. For $3x^2 - 27 = 0$, do not reach for factoring or the formula: isolate the square ($x^2 = 9$) and take roots ($x = \pm 3$) in two steps. The same applies to anything already in squared form, like $(2x - 1)^2 = 16$, where you take roots immediately to get $2x - 1 = \pm 4$. Spotting these forms saves time on the test, because the more general methods would reach the same answer with more work. The cue is simple: if you can get the equation to "(expression)$^2$ = number," use square roots.

Try this

Q1. Solve $(x + 1)^2 = 16$. [1 point]

• Cue. $x + 1 = \pm 4$, so $x = 3$ or $x = -5$.

Q2. Solve $x^2 - 8x + 2 = 0$ by completing the square. [2 points]

• Cue. $x^2 - 8x = -2$; add $16$: $(x - 4)^2 = 14$; $x = 4 \pm \sqrt{14}$.