Complete the square step by step, convert a quadratic to vertex form, and review the vertex, axis of symmetry, discriminant, roots, square term.
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Quadratic equation
Enter coefficients for ax² + bx + c. The result rewrites the quadratic in vertex form and shows the square term used in the completing-the-square step.
1x² − 6x + 5
Examples
Vertex Form
(x − 3)² − 4
The parabola opens up, so the vertex is a minimum. The roots are 2 units left and right of the vertex along the x-axis.
Vertex (h, k)
(3, -4)
Axis of Symmetry
x = 3
Discriminant
16
Square term added
9
Real Roots
1, 5
Square-building check
After factoring out a, the x coefficient inside the parentheses is -6. Half of that is -3, and squaring it gives 9.
Step-by-step
Start with 1x² − 6x + 5
Half of the x coefficient inside: 6 ÷ 2 = 3
Square it: 3² = 9
Complete the square to get vertex form: (x − 3)² − 4
Completing the square calculator: convert quadratics to vertex form with steps
A completing-the-square calculator rewrites a quadratic expression ax² + bx + c in vertex form a(x - h)² + k. This transformation reveals the vertex of the parabola and makes it easier to identify the axis of symmetry, opening direction, minimum or maximum value, discriminant, real roots, and the exact square term used in the algebra.
The vertex form transformation
Starting from ax² + bx + c, factor out a from the first two terms: a(x² + (b/a)x) + c. Then add and subtract (b/(2a))² inside the parentheses to create a perfect square trinomial.
The result is a(x - h)² + k where h = -b/(2a) is the x-coordinate of the vertex and k = c - b²/(4a) is the y-coordinate. If a > 0, the vertex is a minimum; if a < 0, it is a maximum.
h = -b / (2a)
X-coordinate of the vertex. This is the specific relationship the calculator applies when building the result.
k = c - b² / (4a)
Y-coordinate of the vertex. This is the specific relationship the calculator applies when building the result.
Square term = (b / 2a)²
The term added and subtracted inside the completing-the-square step after the leading coefficient is handled.
Why completing the square is useful
Completing the square does more than reformat the quadratic. It turns the equation into a form that immediately exposes the vertex, the axis of symmetry, and whether the parabola opens upward or downward. That makes graphing and optimization questions easier because the most important geometric features are now visible in the formula itself.
The same algebra also explains where the quadratic formula comes from. By moving the constant term, dividing by a, and completing the square, you can isolate x and derive the familiar ±√(b² - 4ac) expression. So this method is both a practical solving tool and a bridge between standard form, vertex form, factoring, and the quadratic formula.
What the calculator adds beyond the final vertex form
Many step-by-step algebra tools show only the transformed expression. This calculator also shows the normalized x coefficient, half of that coefficient, the square term that gets added, the opening direction, the minimum or maximum interpretation, the discriminant, and the distance from the vertex to each real root when real roots exist.
That extra context is useful for learning because completing the square is easy to copy mechanically without understanding the geometry. Seeing the square-building check next to the vertex and root summary connects the algebraic move to the graph.
How to read the result sheet
The vertex form line shows the completed-square expression directly. The vertex row identifies the point (h, k), the axis-of-symmetry row gives the vertical mirror line x = h, and the discriminant helps explain how many real roots the quadratic has before you inspect the solutions.
The step-by-step list is useful when you need to verify each algebra move or study the procedure. If the discriminant is positive, the result sheet will show two real roots. If it is zero, you get one repeated root. If it is negative, the calculator tells you there are no real roots even though the vertex form is still valid.
Worked example: x² - 6x + 5
For x² - 6x + 5, half of the x coefficient is -3, and squaring that gives 9. Rewriting the expression around that perfect square gives (x - 3)² - 4, so the vertex is (3, -4) and the axis of symmetry is x = 3.
From there you can see the parabola opens upward because a = 1 is positive. The discriminant is 16, which confirms there are two real roots, and the calculator returns x = 1 and x = 5. The roots are 2 units to the left and right of the vertex, which is exactly what the completed-square form makes visible.
When a is not 1
The method still works when a is not 1, but the leading coefficient changes the setup. You factor a out of the x² and x terms first, then complete the square inside the parentheses. That is why the square term is based on b/a rather than b alone.
For example, 2x² - 10x - 12 becomes 2(x² - 5x) - 12 before the square is built. Half of -5 is -2.5, and (-2.5)² = 6.25, so the completed-square form keeps the outer factor 2 while adjusting the constant term outside.
It means rewriting part of a quadratic as a perfect square trinomial, then expressing the quadratic in vertex form. For ax² + bx + c, the completed form is a(x - h)² + k.
Why is completing the square useful?
It converts a quadratic to vertex form, instantly revealing the vertex, minimum or maximum point, axis of symmetry, and opening direction. It is also the algebraic basis for deriving the quadratic formula.
What happens when a = 1?
When a = 1, the vertex form simplifies to (x - h)² + k with no leading coefficient. The formulas still apply: h = -b/2 and k = c - b²/4.
Can you complete the square when a is not 1?
Yes. You first factor out a from the x² and x terms, then complete the square inside the parentheses. That is why the method still works for any non-zero quadratic coefficient, not just the simple monic case.
What is the square term you add?
After the x² coefficient inside the parentheses is 1, take half of the x coefficient and square it. In terms of ax² + bx + c, that inside-parentheses square term is (b / 2a)².
Does completing the square solve the quadratic?
It can. Vertex form gives graph information immediately, and if you set the quadratic equal to zero you can use the square-root property to solve for real or complex roots.
What does a negative discriminant mean here?
It means the quadratic has no real x-intercepts, even though the completed-square form still exists. The vertex, axis of symmetry, and direction of opening are still meaningful, but the real-root list will be empty because the parabola never crosses the x-axis.
Is completing the square the same as vertex form?
Completing the square is the algebraic process. Vertex form is the result: a(x - h)² + k. The process creates the form that makes the vertex and axis of symmetry easy to read.
