Completing the Square Calculator

Enter Calculation

a (coefficient of x²)

b (coefficient of x)

c (constant)

Results

Vertex (Completed-Square) Form

1(x − -3)² + -4

a(x − h)² + k

h (horizontal shift)	-3
k (vertical shift)	-4
Vertex (h, k)	(-3, -4)

What is Completing the Square?

Completing the square rewrites a quadratic expression ax² + bx + c into the equivalent vertex form a(x − h)² + k. This form instantly reveals the vertex of the parabola at the point (h, k), making it easy to graph, find maximum or minimum values, and solve quadratic equations.

Completing the square reveals the vertex (h, k) of the parabola.

How to Use the Calculator

Enter the three coefficients: a (the coefficient of x²), b (the coefficient of x), and c (the constant term). The calculator returns the completed-square form along with h, k, and the vertex coordinates. Note that a cannot be zero, otherwise the expression is not quadratic.

The Formula Explained

Starting from ax² + bx + c, factor a from the first two terms and add/subtract the square term. The result is $$ax^2 + bx + c = a\left(x - h\right)^2 + k,\quad h = -\frac{b}{2a},\ k = c - \frac{b^2}{4a}$$ where \(h = -\frac{b}{2a}\) and \(k = c - \frac{b^2}{4a}\). The value h shifts the parabola horizontally and k shifts it vertically.

Geometric meaning: forming a perfect square leaves a small constant piece (k) to add.

Worked Example

For x² + 6x + 5: a = 1, b = 6, c = 5. Then $$h = -\frac{6}{2\cdot 1} = -3 \quad\text{and}\quad k = 5 - \frac{36}{4\cdot 1} = 5 - 9 = -4$$ So x² + 6x + 5 = (x + 3)² − 4, with vertex at (−3, −4).

FAQ

Why is the vertex (h, k)? Because \(a(x - h)^2\) is always ≥ 0 (or ≤ 0 if a is negative), the expression reaches its extreme value of k exactly when x = h.

Does this solve the equation? Setting \(a(x - h)^2 + k = 0\) and solving for x gives the roots, so completing the square is the basis of the quadratic formula.

What if a is negative? The parabola opens downward and the vertex is a maximum, but the same formulas apply.