Completing the Square Calculator

What completing the square does

Completing the square rewrites a quadratic in standard form into an equation that contains a perfect square, so the turning point of the graph can be read straight off the algebra. OpenStax College Algebra works through the same rewrite, and Khan Academy's quadratic unit presents it as a way to solve an equation without reaching for the quadratic formula.

Watch what happens to x^2 + 6x + 5 when the 6 is split into two 3s, giving x^2 + 3x + 3x + 5. Add 9 and subtract 9 so the first two terms can group as x(x + 3) and 3(x + 3). Those two groups share the binomial x + 3, so together they factor to (x + 3)(x + 3), and the leftover 9 - 9 + 5 is a net -4. The equation now reads (x + 3)^2 - 4.

That -4 is not decoration. It is the vertical shift of the parabola, which is why the completed square form hands you the vertex with no extra calculation.

The formula behind the method

Every quadratic can be written as a square plus a remainder. For x^2 + bx + c the recipe is: take half of b, square it, then add and subtract that square so the equality survives.

Half of b is b/2, and the square of that is (b/2)^2. Expanding (x + n)^2 = x^2 + 2nx + n^2 shows why n has to equal half of b, because the middle term 2nx has to reproduce bx. So x^2 + bx + c = (x + b/2)^2 + c - (b/2)^2.

When the leading coefficient is not 1 the same recipe still works, but every term is divided by a first. For ax^2 + bx + c the finished form is a(x + b/2a)^2 + c - b^2/4a, and the turning point sits at -b/2a along the x axis. MIT OpenCourseWare's single variable calculus course describes that same point as the place where a quadratic stops falling and starts rising, which is the geometric meaning of the completed square.

Worked example: x^2 + 6x + 5 = 0

Start with x^2 + 6x + 5 = 0 and move the constant to the right, leaving x^2 + 6x = -5. Half of 6 is 3, and 3^2 = 9, so add 9 to both sides. The left side now contains the square and the right side is 4.

Two quick checks confirm the pair. The sum of the roots should equal -b / a = -6 / 1 = -6, and -1 + (-5) = -6. The product should equal c / a = 5 / 1 = 5, and -1 x (-5) = 5. If either check fails, the arithmetic slipped somewhere in steps 2 through 4.

The vertex comes out of the same work. The square (x + 3)^2 is zero at x = -3, and there the whole expression sits at -4, so the vertex is (-3, -4) and the parabola opens upward with a minimum of -4.

Worked example: 2x^2 - 8x + 3 = 0

A leading coefficient that is not 1 adds one step at the front. Divide every term by 2 to get x^2 - 4x + 1.5 = 0, and from there the method is identical. Half of -4 is -2, and squaring -2 gives 4.

The rounded roots satisfy the equation to within a rounding error, which is the expected behavior at four decimal places. The vertex now sits at (2, -5), because the square (x - 2)^2 vanishes at x = 2 and the leftover constant there is -5. A quick check on the sum of the roots: 3.5811 + 0.4189 = 4, and -b / a = 8 / 2 = 4.

Multiplying the finished form back by 2 gives the vertex form 2(x - 2)^2 - 5, which is the shape you would hand to a graphing tool or a spreadsheet.

Vertex form and what the graph does

Vertex form is a(x - h)^2 + k, where the vertex sits at (h, k). The completed square puts an equation into that shape, so the graph needs no plotting. If a is positive the parabola opens upward and k is the minimum value of the function. If a is negative it opens downward and k is the maximum.

The vertex coordinates also come from the original coefficients without doing the rewrite. The x coordinate is -b/2a, so for x^2 + 6x + 5 it is -6 / 2 = -3, and the y coordinate is c - b^2/4a, which is 5 - 36 / 4 = 5 - 9 = -4. For 2x^2 - 8x + 3 the same pair reads 8 / 4 = 2 and 3 - 64 / 8 = 3 - 8 = -5.

This is where completing the square earns its keep. A school assignment that asks for the vertex, the minimum value or the range of a quadratic can be answered with the formula, while the quadratic formula would give the roots and leave the graph work still ahead.

Completing the square against the quadratic formula

The two methods agree on the roots, because the quadratic formula is nothing more than the completed square rearranged. Starting from (x + b/2a)^2 = (b^2 - 4ac)/4a^2 and taking square roots on both sides leaves x = -b/2a +/- sqrt(b^2 - 4ac)/2a, which is the familiar formula in a different costume.

Reach for factoring first when the numbers cooperate, since x^2 + 6x + 5 factors as (x + 1)(x + 5) and finishes in seconds. Reach for completing the square when factoring stalls, when a vertex is needed, or when a task asks for standard form, vertex form or a graph. Reach for the quadratic formula when speed matters and only the roots are wanted, for example x^2 + 6x + 3, which does not factor over the integers and gives x = -3 +/- sqrt(6), roughly -0.5505 and -5.4495.

What the calculator returns

The defaults are the first worked example: a = 1, b = 6 and c = 5, which is the equation x^2 + 6x + 5.

Type 2, -8 and 3 into the three boxes and the same panel reports 2(x - 2)^2 - 5 = 0, the square root step (x - 2)^2 = 2.5, the vertex (2, -5), a discriminant of 40 and the roots x = 3.5811 and x = 0.4189.

Two edge cases are handled explicitly. A leading coefficient of 0 is not a quadratic, so the tool asks for a different a instead of dividing by zero. A negative discriminant produces complex roots, reported as x = -1 + 2i and x = -1 - 2i for x^2 + 2x + 5, which is the same answer the quadratic formula gives.

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