Completing the Square Calculator

Completing the Square Guide

What's the key thing to understand about The Method for a = 1?

For x² + bx + c: write as (x + b/2)² − (b/2)² + c. Example: x² + 6x + 7. Half of 6 is 3. (x+3)² − 9 + 7 = (x+3)² − 2. Vertex form: (x+3)² − 2. Turning point: (−3, −2). The x-coordinate of the vertex is always −(b/2a). The y-coordinate is c − b²/(4a). Check: expand (x+3)² − 2 = x² + 6x + 9 − 2 = x² + 6x + 7 ✓.

What do I need to know about Completing the Square for a ≠ 1?

For ax² + bx + c: factor out a first. 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2(x+2)² − 8 + 3 = 2(x+2)² − 5. Turning point: (−2, −5). Alternatively: use the formula. Vertex x-coordinate: −b/(2a) = −8/(2×2) = −2. Vertex y-coordinate: substitute x = −2 into the original: 2(4) + 8(−2) + 3 = 8 − 16 + 3 = −5. The turning point is (−2, −5).

What do I need to know about Solving Quadratics by Completing the Square?

From vertex form (x+p)² = q, take square roots: x + p = ±√q. x = −p ± √q. Example: x² + 6x + 7 = 0 → (x+3)² = 2 → x = −3 ± √2 = −3 + 1.414... = −1.586 or −3 − 1.414 = −4.414. If q < 0: no real roots (parabola does not cross x-axis). If q = 0: repeated root (parabola just touches x-axis). If q > 0: two distinct roots. Completing the square derives the quadratic formula: x = (−b ± √(b²−4ac)) / 2a.

What's the key thing to understand about Applications of the Vertex Form?

The vertex form y = a(x−h)² + k immediately shows: turning point (h, k). Whether minimum (a > 0) or maximum (a < 0). The axis of symmetry x = h. The minimum/maximum value (k). This form is essential in physics: projectile motion (height vs time is a negative quadratic — vertex gives maximum height). Economics: profit maximisation (revenue minus cost is often a downward quadratic — vertex gives the profit-maximising quantity). Engineering: cable suspension bridge profiles follow parabolic, quadratic curves, and completing the square helps engineers find the exact lowest point of the cable's sag.

What's the key thing to understand about the Discriminant in Vertex Form?

Once a quadratic is written in vertex form a(x−h)² + k, the sign of k (relative to the parabola's orientation) tells you immediately how many real roots exist — without needing a separate discriminant calculation. If a > 0 (parabola opens upward) and k < 0, the vertex sits below the x-axis, so the curve must cross it twice: two real roots. If k = 0, the vertex touches the x-axis exactly: one repeated root (the discriminant b²−4ac = 0). If k > 0 with a > 0, the whole parabola sits above the x-axis: no real roots. For x² − 6x + 10, completing the square gives (x−3)² + 1 — vertex (3, 1), and since a=1>0 and k=1>0, there are no real roots, confirming b²−4ac = 36−40 = −4 < 0. This is a quick way to sanity-check a completed-square result against the discriminant you'd get from the quadratic formula.

What should I know about Completing the Square for Circles?

The same technique used for quadratics converts a circle's general equation x² + y² + Dx + Ey + F = 0 into centre-radius form (x−a)² + (y−b)² = r². Complete the square separately on the x-terms and the y-terms. Example: x² + y² − 6x + 4y − 12 = 0. Group: (x² − 6x) + (y² + 4y) = 12. Complete each square: (x−3)² − 9 + (y+2)² − 4 = 12, so (x−3)² + (y+2)² = 25. This gives centre (3, −2) and radius 5 (since 25 = 5²). If the constant on the right ends up negative, the 'circle' has no real points — it's an imaginary locus, a useful check when verifying a derived circle equation. This cross-over is why the completing-the-square method taught for quadratics transfers directly to conic sections, which is why you'll see it again on this site's circle equation calculator.

What should I know about Why Completing the Square Derives the Quadratic Formula?

The quadratic formula itself is derived by completing the square on the general equation ax² + bx + c = 0. Dividing through by a gives x² + (b/a)x + c/a = 0; completing the square on the x terms and rearranging eventually produces x = [-b ± √(b² - 4ac)] / 2a — the familiar formula. This means the quadratic formula isn't a separate rule to memorize independently of completing the square; it's simply that process carried out once, in general, so it never has to be repeated for a specific equation again. Seeing this derivation clarifies why the discriminant, b² - 4ac, appears under a square root in the formula — it's a direct leftover of the "add and subtract (b/2a)²" step that completing the square always requires.

What should I know about Completing the Square for Integration?

Completing the square has an important use in calculus beyond solving quadratic equations: it is often the key step for integrating rational functions whose denominator is a quadratic that does not factor neatly into real linear terms. An integral such as ∫ 1/(x² + 4x + 13) dx cannot be evaluated using simple algebraic substitution directly, because the quadratic denominator x² + 4x + 13 has a negative discriminant (16 − 52 = −36) and therefore no real roots, so it cannot be split into partial fractions over the real numbers. Completing the square rewrites the denominator as (x+2)² + 9, revealing the integral's true structure: a shifted variable u = x+2 raised to a power plus a constant, matching the standard integral form ∫ 1/(u² + a²) du = (1/a)arctan(u/a) + C. Substituting back gives the final result (1/3)arctan((x+2)/3) + C. This same technique applies to a wide family of integrals involving arctangent, and to related forms involving arcsine, whenever the integrand contains an irreducible quadratic expression in the denominator or under a square root — making completing the square not just an algebra technique for finding vertices, but a genuine calculus tool for unlocking integrals that otherwise resist direct methods.

Completing the Square: Quick Reference Table

For a quadratic in the form ax² + bx + c, completing the square converts it to vertex form a(x − h)² + k, where the vertex coordinates follow directly from a, b, and c:

Once in vertex form, you can read off the maximum or minimum value of the quadratic without calculus — it's simply k, occurring at x = h.

Worked Example: Completing the Square to Find the Maximum Height of a Projectile

A ball's height in meters t seconds after being thrown is modeled by h(t) = −5t² + 20t + 1. Find the maximum height it reaches and when. Step 1 — factor out the leading coefficient from the first two terms: h(t) = −5(t² − 4t) + 1. Step 2 — complete the square inside the parentheses: half of −4 is −2, squared is 4, so t² − 4t = (t − 2)² − 4. Step 3 — substitute back and distribute the −5: h(t) = −5[(t − 2)² − 4] + 1 = −5(t − 2)² + 20 + 1 = −5(t − 2)² + 21. The vertex form shows the maximum value is 21 (meters), reached at t = 2 (seconds), since −5(t−2)² can never be positive — it's zero at t = 2 and negative everywhere else, meaning h(t) can never exceed 21. Checking directly: h(2) = −5(4) + 20(2) + 1 = −20 + 40 + 1 = 21 ✓. This is exactly how physics and engineering problems use completing the square to find a maximum or minimum without needing calculus — the same technique that would otherwise require setting a derivative to zero.

Completing the Square with Negative or Non-Integer Leading Coefficients

The method works the same way even when a is negative or fractional — you just need to be extra careful with signs when factoring a out of the first two terms. Complete the square for f(x) = −2x² + 8x − 3. Factor −2 out of the first two terms only: f(x) = −2(x² − 4x) − 3. Complete the square inside the parentheses: half of −4 is −2, squared is 4, so x² − 4x = (x−2)² − 4. Substitute back: f(x) = −2[(x−2)² − 4] − 3 = −2(x−2)² + 8 − 3 = −2(x−2)² + 5. So the vertex is at (2, 5), and since a = −2 is negative, this is a maximum value of 5. Check by direct substitution: f(2) = −2(4) + 8(2) − 3 = −8 + 16 − 3 = 5 ✓. The negative leading coefficient flips the parabola to open downward, which is why the vertex represents a maximum rather than a minimum here, exactly mirroring how a positive a produces a minimum.