upgrade
upgrade

📘Intermediate Algebra

Completing the Square Method

Study smarter with Fiveable

Get study guides, practice questions, and cheatsheets for all your subjects. Join 500,000+ students with a 96% pass rate.

Get Started

Why This Matters

Completing the square isn't just another way to solve quadratic equations—it's the foundation for understanding why the quadratic formula works, how parabolas shift on a coordinate plane, and what vertex form actually reveals about a function. When you're tested on transforming quadratics, deriving the quadratic formula, or analyzing the vertex of a parabola, completing the square is the underlying skill that makes it all click.

This method transforms any quadratic from standard form into a perfect square structure, giving you direct access to solutions and graph behavior. You're being tested on your ability to manipulate algebraic expressions, maintain equation balance, and connect algebraic steps to geometric meaning. Don't just memorize the procedure—understand what each step accomplishes and why the algebra works.


Setting Up the Equation

Before you can complete the square, you need the equation in the right form. These initial steps ensure you're working with a structure that can become a perfect square trinomial.

Identify Standard Form

  • Standard form is ax2+bx+c=0ax^2 + bx + c = 0—confirm your equation matches this structure before proceeding
  • The coefficient aa cannot equal zero; if it does, you have a linear equation, not a quadratic
  • Label your coefficients clearly (aa, bb, and cc) since you'll reference them throughout the process

Isolate the Variable Terms

  • Move the constant cc to the right side—this creates space on the left for your perfect square
  • Change the sign when moving terms across the equals sign to maintain balance
  • Your equation should now read ax2+bx=cax^2 + bx = -c, with all variable terms on the left

Normalize the Leading Coefficient

  • Factor out aa if it's not equal to 1—completing the square requires a leading coefficient of 1
  • Divide the entire equation by aa as an alternative approach that achieves the same result
  • Adjust the right side accordingly; if you factor out aa from the left, the right becomes ca\frac{-c}{a}

Compare: Factoring out aa vs. dividing by aa—both normalize the leading coefficient, but factoring keeps you working with integers longer while dividing may introduce fractions earlier. Choose based on the numbers you're given.


Creating the Perfect Square

This is the heart of the method. You're engineering a perfect square trinomial by strategically adding the same value to both sides. The key insight: a perfect square trinomial always equals the square of a binomial.

Calculate the Completing Term

  • Take half of the xx-coefficient, then square it—this value, (b2)2\left(\frac{b}{2}\right)^2, is what "completes" the square
  • This works because (x+d)2=x2+2dx+d2(x + d)^2 = x^2 + 2dx + d^2, so d=b2d = \frac{b}{2} makes the middle term match
  • Add this value to both sides to maintain equation balance—never add to just one side

Factor the Left Side

  • Rewrite the trinomial as (x+b2)2\left(x + \frac{b}{2}\right)^2—this is the payoff of completing the square
  • The sign inside the binomial matches the sign of bb; if bb was negative, you'll have (xd)2(x - d)^2
  • Verify by expanding your binomial to confirm it produces the original trinomial

Compare: Perfect square trinomials vs. factorable trinomials—not every trinomial factors nicely, but any quadratic can be completed into a perfect square. This is why completing the square is more universally applicable than simple factoring.


Solving for the Variable

With your equation now in the form (x+b2)2=k\left(x + \frac{b}{2}\right)^2 = k, you can extract the solutions using inverse operations.

Apply the Square Root Property

  • Take the square root of both sides, which "undoes" the squaring operation
  • Include ±\pm on the right side—every positive number has two square roots, and both are valid solutions
  • Your equation becomes x+b2=±kx + \frac{b}{2} = \pm\sqrt{k}, where kk is whatever value accumulated on the right

Isolate xx

  • Subtract b2\frac{b}{2} from both sides to get xx alone
  • Write both solutions explicitly: x=b2+kx = -\frac{b}{2} + \sqrt{k} and x=b2kx = -\frac{b}{2} - \sqrt{k}
  • Simplify radicals when possible—look for perfect square factors under the radical

Compare: Completing the square vs. the quadratic formula—the quadratic formula is actually derived from completing the square on ax2+bx+c=0ax^2 + bx + c = 0. Understanding this connection helps you remember the formula and know when each method is more efficient.


Verification and Interpretation

Solving isn't complete until you've confirmed your answers and understood what they mean.

Check Your Solutions

  • Substitute each solution back into the original equation—both sides should equal the same value
  • Common errors include sign mistakes and arithmetic errors when calculating the completing term
  • If solutions don't check, trace back to the step where you calculated (b2)2\left(\frac{b}{2}\right)^2—this is where most mistakes occur

Connect to Graphical Meaning

  • The solutions are the xx-intercepts of the parabola y=ax2+bx+cy = ax^2 + bx + c
  • Vertex form y=a(xh)2+ky = a(x - h)^2 + k comes directly from completing the square—the vertex is (h,k)(h, k)
  • If the right side becomes negative, there are no real solutions; the parabola doesn't cross the xx-axis

Compare: Real solutions vs. no real solutions—when k<0k < 0 after completing the square, you'd need the square root of a negative number. This signals the parabola sits entirely above or below the xx-axis, depending on the sign of aa.


Quick Reference Table

ConceptKey Steps or Ideas
Standard Form SetupIdentify aa, bb, cc; move constant to right side
NormalizingFactor out or divide by aa when a1a \neq 1
Completing TermCalculate (b2)2\left(\frac{b}{2}\right)^2; add to both sides
Perfect Square StructureLeft side becomes (x+d)2(x + d)^2 where d=b2d = \frac{b}{2}
Square Root PropertyTake ±both sides\pm\sqrt{\text{both sides}}; remember both roots
Solution IsolationSolve for xx; write both explicit solutions
VerificationSubstitute solutions into original equation
Graphical ConnectionSolutions = xx-intercepts; completed form = vertex form

Self-Check Questions

  1. Why must you add the completing term to both sides of the equation, and what would happen if you only added it to the left?

  2. Given 2x2+8x10=02x^2 + 8x - 10 = 0, what value do you add to both sides after normalizing the leading coefficient, and why?

  3. Compare solving x2+6x+5=0x^2 + 6x + 5 = 0 by factoring versus completing the square—which is more efficient here, and when would completing the square be the better choice?

  4. If completing the square on x24x+7=0x^2 - 4x + 7 = 0 results in (x2)2=3(x - 2)^2 = -3, what does this tell you about the graph of y=x24x+7y = x^2 - 4x + 7?

  5. How does the completed square form of a quadratic relate to vertex form, and what information can you read directly from it without further calculation?