Solve quadratic equations by taking square roots and by completing the square, and use completing the square to rewrite a quadratic in vertex form (Ohio A-REI.4a, A-REI.4b, F-IF.8).

What this topic is asking

Ohio standard A-REI.4a asks you to solve quadratics by taking square roots and by completing the square, and to use completing the square to rewrite a quadratic in vertex form. The square-root method handles equations with no linear term; completing the square handles any quadratic and reveals the vertex. Vertex form is not on the reference sheet. These methods sit in the Quadratics block.

The square-root method

If the variable appears only squared (no separate $x$ term), isolate the square and root both sides.

The plus-or-minus is the heart of the method: a positive number has two square roots, and dropping the negative one loses a solution.

Completing the square

Completing the square forces a perfect-square trinomial so you can finish with the square-root method, and it works for any quadratic.

Producing vertex form

Completing the square also rewrites $y = x^2 + bx + c$ as vertex form $y = (x - h)^2 + k$, where $(h, k)$ is the vertex. For $y = x^2 + 6x - 7$, completing the square gives $y = (x + 3)^2 - 16$, so the vertex is $(-3, -16)$. This is why the method is valued: it both solves and reveals the parabola's turning point.

How Ohio examines this topic

• Equation and numeric response. Solve a square-root-form equation, or the value that completes a square.
• Multiple choice and multiple-select. Pick the completing value, the solutions, or the vertex form.
• Drag and drop. Order the steps of completing the square.

Why the plus-or-minus is required

When you take the square root as a step in solving an equation, you must allow both signs because two different numbers square to the same positive value: $4^2 = 16$ and $(-4)^2 = 16$. Writing only $x = \sqrt{16} = 4$ answers "what is the principal square root?" but the equation $x^2 = 16$ asks "what values of $x$ make this true?", and both $4$ and $-4$ do. So the $\pm$ is not optional decoration, it captures the second solution. Forgetting it is the single most common way to lose a root with the square-root method, and it is exactly the kind of slip an exact-match item penalizes.

Why half-the-coefficient-squared works

Completing the square reverses the expansion of a perfect square. Expanding $(x + p)^2$ gives $x^2 + 2px + p^2$, so the middle coefficient is $2p$ and the constant is $p^2$. Reading this backward: if the middle coefficient is $b = 2p$, then $p = \frac{b}{2}$, and the constant needed to complete the square is $p^2 = \left(\frac{b}{2}\right)^2$. That is why the rule is "half the middle coefficient, squared." Adding exactly that amount manufactures the perfect-square trinomial $(x + \frac{b}{2})^2$, which factors cleanly and lets you finish by taking square roots. Understanding the connection to $(x + p)^2$ makes the rule something you can reconstruct rather than memorize blindly.

Try this

Q1. Solve $x^2 = 49$. [1 point]

• Cue. $x = \pm 7$.

Q2. What completes the square for $x^2 - 10x$? [1 point]

• Cue. $\left(\frac{-10}{2}\right)^2 = 25$.