Algebra Bridge

Completing the Square

Completing the square rewrites a quadratic expression as a perfect square plus or minus a constant. It is the bridge from quadratic formulas to vertex form, circles, and conic sections.

Solved Examples

Jump directly to a worked example by problem type.

Fact Table

Goal Move Result
Turn $x^2+bx$ into a square. Add $\left(\frac b2\right)^2$. $x^2+bx+\left(\frac b2\right)^2=\left(x+\frac b2\right)^2$
Leading coefficient is not 1. Factor $a$ from the quadratic and linear terms first. Complete the square inside parentheses.
Equation must stay balanced. Add the same value to both sides. No solution set changes.
Expression is being rewritten. Add and subtract the same value. The expression stays equivalent.

Content Formulas

Square Add-On
$$\left(\frac b2\right)^2$$
Perfect Square
$$x^2+bx+\left(\frac b2\right)^2=\left(x+\frac b2\right)^2$$
For conics, group x-terms together and y-terms together before completing each square.

Classic Examples

Rewrite in Vertex Form

Write $y=x^2+6x+5$ in vertex form.

Solution Steps
  1. Write the original quadratic expression.
  2. Add and subtract the square of half the linear coefficient.
  3. Rewrite the perfect-square trinomial.
  4. State the vertex: $(-3,-4)$.
$$y=x^2+6x+5$$ $$y=x^2+6x+9-9+5$$ $$y=(x+3)^2-4$$ $$\text{vertex: }(-3,-4)$$

Leading Coefficient

Write $y=2x^2-12x+10$ in vertex form.

Solution Steps
  1. Factor the leading coefficient from the quadratic and linear terms.
  2. Add and subtract the square of half the linear coefficient inside the parentheses.
  3. Rewrite the perfect-square trinomial and distribute the leading coefficient.
  4. State the vertex form: $y=2(x-3)^2-8$.
$$y=2(x^2-6x)+10$$ $$y=2(x^2-6x+9-9)+10$$ $$y=2(x-3)^2-18+10$$ $$y=2(x-3)^2-8$$