Completing the Square Calculator

Rewrite a quadratic in vertex form and solve it

Completing the Square Calculator — free online tool

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Working out how to rewrite a quadratic in vertex form and solve it is easier with Completing the Square Calculator, which answers as you type. Type Coefficient of x-squared, Coefficient of x and Constant term into Completing the Square; each keystroke re-runs the math. In Completing the Square the working sits beside the answer. Use Completing the Square for a reliable number, fast. Nothing is uploaded while Completing the Square works, and the figures stay put. Completing the Square is part of the Math collection on CalcProMaster, alongside completing the square calculator vertex form quadratic solver, free online completing the square calculator and more. No sign-up or email — Completing the Square answers as you type.

How do you complete the square?

Completing the square rewrites a quadratic as a perfect square plus a constant, which is what makes the vertex and the roots readable. Enter the coefficients and the page shows each step, then reports the vertex form, the vertex itself and the discriminant.

Half of 6 is 3, and squaring it gives 9 — the number added and then subtracted to keep the expression unchanged.

How the rewrite works

Take a(x + b/2a)² + (c − b²/4a). For x² + 6x + 5, b/2a is 3, so the bracket is (x + 3)², which expands to x² + 6x + 9 — four more than the original. Subtracting 4 corrects it. That remainder is k, the vertical offset, and k = c − b²/4a = 5 − 9 = −4.

The vertex is (−b/2a, k). Because a is positive the parabola opens upward, so the vertex is the minimum: the expression can never go below −4.

Worked examples

x² + 6x + 5 → (x + 3)² − 4, roots −1 and −5
0.5x² + 6x + 5 → 0.5(x + 6)² − 13, roots −0.90 and −11.10
2x² + 6x + 5 → 2(x − 1.5)² + 0.5, no real roots
x² + 12x + 5 → (x + 6)² − 31, roots −0.43 and −11.57
x² + 6x + 2.5 → (x + 3)² − 6.5, roots −0.45 and −5.55
x² + 3x + 5 → (x + 1.5)² + 2.75, no real roots

Three of those six never cross zero, and in each case k is positive — the whole parabola lives above the axis. Doubling a from 1 to 2 makes the parabola narrower and lifts the vertex to y = 0.5, which is enough to eliminate the roots entirely even though b and c did not change.

Why the form is more useful than the roots

Vertex form answers questions the quadratic formula does not. The minimum of x² + 6x + 5 is −4, reached at x = −3, which is what you need for optimisation, for sketching, and for reading a parabola off a graph. The roots only tell you where the curve meets the axis.

The discriminant, b² − 4ac, decides how many real solutions exist before any square root is taken. Here it is 36 − 20 = 16, a perfect square, which is why both roots are whole numbers. When it is negative the rewrite still works — only the last step of solving fails, because a squared quantity cannot be negative.

One caveat: a must not be zero. Dividing b by 2a is the first operation, and the page returns an error rather than inventing a value.

Frequently Asked Questions

What is x² + 6x + 5 in vertex form?

(x + 3)² − 4. The vertex is at (−3, −4), and because a is positive that point is the minimum of the curve.

How do you find the roots from vertex form?

Set the square equal to −k/a. Here (x + 3)² = 4, so x + 3 = ±2, giving x = −1 and x = −5.

What does a negative discriminant mean here?

The vertex sits above the x-axis and the parabola never reaches zero, so there are no real roots. The tool says so instead of printing an error.

Why is the number 9 subtracted?

Squaring (x + 3) adds 9 that was not in the original expression. Subtracting b²/4a keeps the value identical while making the square visible.

Does the method work when a is not 1?

Yes. Factor a out of the first two terms first, or use b/2a and c − b²/4a directly as this page does; 0.5x² + 6x + 5 becomes 0.5(x + 6)² − 13.

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