{"id":2440,"date":"2026-08-12T11:59:50","date_gmt":"2026-08-12T11:59:50","guid":{"rendered":"https:\/\/www.think10x.ai\/blog\/?p=2440"},"modified":"2026-08-12T12:06:28","modified_gmt":"2026-08-12T12:06:28","slug":"how-to-complete-the-square-step-by-step","status":"publish","type":"post","link":"https:\/\/www.think10x.ai\/blog\/how-to-complete-the-square-step-by-step\/","title":{"rendered":"How to Complete the Square Step by Step"},"content":{"rendered":"\t\t
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To complete the square, move the constant to one side of the equation, take half the coefficient of x, square it, and add that value to both sides. This turns the left side into a perfect square binomial, which you can solve by taking the square root of both sides. It works on every quadratic equation, even ones that cannot be factored.<\/span><\/p>\n

What Does It Mean to Complete the Square?<\/h2>\n

Completing the square is a method for rewriting a quadratic expression so it contains a perfect square binomial instead of two separate x-terms. A quadratic like x\u00b2 + 6x is not a perfect square on its own, but adding 9 turns it into x\u00b2 + 6x + 9, which factors neatly as (x + 3)\u00b2.<\/span><\/p>\n

Beyond solving for x, this same technique shows up when graphing a parabola or finding its highest or lowest point, since vertex form puts that information directly in the equation. It also underlies the derivation of the quadratic formula, so a solid grasp of completing the square makes most other quadratic topics easier to follow.<\/span><\/p>\n

The Completing the Square Formula<\/h2>\n

The completing the square formula starts from the standard quadratic form:<\/span><\/p>\n

ax\u00b2 + bx + c = 0<\/b><\/p>\n

To complete the square, rearrange it into:<\/span><\/p>\n

a(x + b\/2a)\u00b2 = (b\u00b2 \u2212 4ac) \/ 4a<\/b><\/p>\n

The key number to remember is (b\/2)\u00b2, or half the coefficient of x, squared. This is the value that turns x\u00b2 + bx into a perfect square trinomial. In fact, the Common Core State Standards for high school algebra specifically require students to use this method to transform any quadratic equation into the form (x \u2212 p)\u00b2 = q, and then derive the quadratic formula from that same transformation.\u00a0<\/span><\/p>\n

A quick way to check your work is to expand the result back out. If a(x + b\/2a)\u00b2 simplifies back to the original ax\u00b2 + bx + c, you completed the square correctly. This habit catches the most common error in this method: adding the constant to only one side instead of both.<\/span><\/p>\n

How to Complete the Square Step by Step<\/h2>\n

Here is how to complete the square step by step, using x\u00b2 + 8x + 7 = 0 as a working example.<\/span><\/p>\n

    \n
  1. Move the constant to the other side.<\/b> Subtract 7 from both sides: x\u00b2 + 8x = \u22127.<\/span><\/li>\n
  2. Take half the coefficient of x, then square it.<\/b> Half of 8 is 4, and 4\u00b2 = 16.<\/span><\/li>\n
  3. Add that value to both sides.<\/b> x\u00b2 + 8x + 16 = \u22127 + 16, which simplifies to x\u00b2 + 8x + 16 = 9.<\/span><\/li>\n
  4. Factor the left side as a perfect square.<\/b> x\u00b2 + 8x + 16 factors into (x + 4)\u00b2.<\/span><\/li>\n
  5. Take the square root of both sides.<\/b> (x + 4)\u00b2 = 9 becomes x + 4 = \u00b13.<\/span><\/li>\n
  6. Solve for x.<\/b> x = \u22124 + 3 = \u22121, or x = \u22124 \u2212 3 = \u22127.<\/span><\/li>\n<\/ol>\n

    When the coefficient of x\u00b2 is not 1<\/h3>\n

    If the equation looks like 2x\u00b2 + 12x + 10 = 0, divide every term by the leading coefficient first. Dividing by 2 gives x\u00b2 + 6x + 5 = 0, and from there, the same six steps apply exactly as before. This division step is the one most students forget, so check for a leading coefficient before doing anything else. Seeing this complete-the-square method applied to a few different coefficients in a<\/span> step-by-step video explanation<\/span><\/a> makes the pattern automatic faster than the rule alone.<\/span><\/p>\n

    Complete the Square Method: Worked Examples<\/h2>\n

    Example 1. Solving a quadratic equation<\/h3>\n

    Solve x\u00b2 \u2212 4x \u2212 5 = 0 using the complete-the-square method.<\/span><\/p>\n

    Move the constant: x\u00b2 \u2212 4x = 5. Half of \u22124 is \u22122, and (\u22122)\u00b2 = 4. Add 4 to both sides: x\u00b2 \u2212 4x + 4 = 9. Factor: (x \u2212 2)\u00b2 = 9. Take the square root: x \u2212 2 = \u00b13. Solve: x = 5 or x = \u22121.<\/span><\/p>\n

    Example 2. A coefficient other than 1<\/h3>\n

    Solve 3x\u00b2 + 12x \u2212 15 = 0.<\/span><\/p>\n

    Divide everything by 3: x\u00b2 + 4x \u2212 5 = 0. Move the constant: x\u00b2 + 4x = 5. Half of 4 is 2, and 2\u00b2 = 4. Add 4 to both sides: x\u00b2 + 4x + 4 = 9. Factor: (x + 2)\u00b2 = 9. Take the square root: x + 2 = \u00b13. Solve: x = 1 or x = \u22125.<\/span><\/p>\n

    Example 3. Converting to vertex form<\/h3>\n

    Rewrite f(x) = x\u00b2 \u2212 6x + 2 in vertex form.<\/span><\/p>\n

    Half of \u22126 is \u22123, and (\u22123)\u00b2 = 9. Rewrite as x\u00b2 \u2212 6x + 9 \u2212 9 + 2, which groups into (x \u2212 3)\u00b2 \u2212 7. The vertex form is f(x) = (x \u2212 3)\u00b2 \u2212 7, showing the parabola’s vertex sits at (3, \u22127). Watching this rearrangement happen one line at a time in a<\/span> narrated video walkthrough<\/span><\/a> makes the regrouping step click faster than reading it cold.<\/span><\/p>\n

    Example 4. A word problem using vertex form<\/h3>\n

    A ball is thrown so its height in meters is modeled by h(t) = \u22125t\u00b2 + 20t, where t is time in seconds. Find the maximum height.<\/span><\/p>\n

    Factor out \u22125: h(t) = \u22125(t\u00b2 \u2212 4t). Half of \u22124 is \u22122, and (\u22122)\u00b2 = 4. Add and subtract inside the parentheses: h(t) = \u22125(t \u2212 2)\u00b2 + 20. The vertex is at (2, 20), so the ball reaches a maximum height of 20 meters at t = 2 seconds. This is where completing the square beats factoring, since the vertex gives the maximum directly.<\/span><\/p>\n

    Completing the Square Practice Problems<\/h2>\n

    Try these <\/span>completing-the-square practice problems<\/b>, then check your work against the solutions below.<\/span><\/p>\n

    Problem 1.<\/b>\u00a0Solve x\u00b2 + 10x + 21 = 0. Solution: x\u00b2 + 10x = \u221221. Half of 10 is 5, and 5\u00b2 = 25. x\u00b2 + 10x + 25 = 4. Factor: (x + 5)\u00b2 = 4. So x + 5 = \u00b12, giving x = \u22123 or x = \u22127.<\/span><\/p>\n

    Problem 2.<\/b>\u00a0Solve 2x\u00b2 \u2212 8x \u2212 24 = 0. Solution: Divide by 2: x\u00b2 \u2212 4x \u2212 12 = 0. Move the constant: x\u00b2 \u2212 4x = 12. Half of \u22124 is \u22122, and (\u22122)\u00b2 = 4. x\u00b2 \u2212 4x + 4 = 16. Factor: (x \u2212 2)\u00b2 = 16. So x \u2212 2 = \u00b14, giving x = 6 or x = \u22122.<\/span><\/p>\n

    Problem 3.<\/b>\u00a0Convert f(x) = x\u00b2 + 2x \u2212 8 to vertex form. Solution: Half of 2 is 1, and 1\u00b2 = 1. Rewrite as x\u00b2 + 2x + 1 \u2212 1 \u2212 8, which groups into (x + 1)\u00b2 \u2212 9.<\/span><\/p>\n

    Checklist before you submit an answer<\/b><\/p>\n