{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe complete the square and then factor a difference of two squares.\n\n1. Let $a, b, c \\in \\C$ with $a \\ne 0$, and let $s \\in \\C$ satisfy $s^2 = b^2 - 4ac$. Fix $x \\in \\C$.\n2. The [[complex numbers|09FTAET5Z4]] [[form a field|0FRSR10DWF]]. So the usual rules of algebra hold, and each nonzero number has an inverse.\n3. Expanding gives $(2ax + b)^2 = 4a^2x^2 + 4abx + b^2$.\n4. By Step 3, $4a(ax^2 + bx + c) = (2ax + b)^2 - (b^2 - 4ac)$.\n5. Since $s^2 = b^2 - 4ac$, Step 4 becomes $4a(ax^2 + bx + c) = (2ax + b)^2 - s^2$.\n6. For all $u, v \\in \\C$ we have $u^2 - v^2 = (u - v)(u + v)$, by expanding the right side.\n7. By Steps 5 and 6, $4a(ax^2 + bx + c) = (2ax + b - s)(2ax + b + s)$.\n8. If $u, v \\in \\C$ and $uv = 0$, then $u = 0$ or $v = 0$. Indeed, if $u \\ne 0$, then $v = u^{-1}(uv) = u^{-1} \\cdot 0 = 0$.\n9. The number $4a$ is nonzero, since $4 \\ne 0$ and $a \\ne 0$, by Step 8. Likewise $2a \\ne 0$.\n10. If $ax^2 + bx + c = 0$, then $4a(ax^2 + bx + c) = 0$. Conversely, if $4a(ax^2 + bx + c) = 0$, then $ax^2 + bx + c = 0$ by Steps 8 and 9. So the two equations are equivalent.\n11. By Steps 7, 8 and 10, $ax^2 + bx + c = 0$ holds if and only if $2ax + b - s = 0$ or $2ax + b + s = 0$.\n12. Since $2a \\ne 0$, the equation $2ax + b - s = 0$ is equivalent to $x = \\frac{-b + s}{2a}$. We add $s - b$ to both sides and multiply by $(2a)^{-1}$.\n13. In the same way, $2ax + b + s = 0$ is equivalent to $x = \\frac{-b - s}{2a}$.\n14. By Steps 11, 12 and 13, $ax^2 + bx + c = 0$ if and only if $x = \\frac{-b + s}{2a}$ or $x = \\frac{-b - s}{2a}$.\n\n## Idea\n\nMultiplying the equation by $4a$ turns its left side into the perfect square $(2ax + b)^2$ minus the discriminant.\nSince the discriminant is $s^2$, the result is a difference of two squares, which factors into two linear factors.\nA product in a field is zero only when a factor is zero, so the roots are the zeros of the two linear factors.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "12D10"
    ],
    "name": "Proof of the quadratic formula by completing the square",
    "proves": "0KYZXYDGQZ",
    "sources": [
      {
        "locator": "Derivation by completing the square",
        "url": "https://en.wikipedia.org/wiki/Quadratic_formula"
      }
    ],
    "tag": "NXNBXZWFZ1",
    "uses": [
      "0FRSR10DWF",
      "09FTAET5Z4"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/NXNBXZWFZ1/",
  "links": [
    {
      "field": "proves",
      "target": "0KYZXYDGQZ"
    },
    {
      "field": "uses",
      "target": "0FRSR10DWF"
    },
    {
      "field": "uses",
      "target": "09FTAET5Z4"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/NXNBXZWFZ1.md",
  "msc": [
    "12D10"
  ],
  "name": "Proof of the quadratic formula by completing the square",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/NXN/NXNBXZWFZ1.md",
  "status": null,
  "tag": "NXNBXZWFZ1",
  "uri": "m2://NXNBXZWFZ1",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-quadratic-formula-by-completing-the-square/",
  "writing": {
    "tag": "NXNBXZWFZ1",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}