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Rewriting Expressions (Completing the Square)

  • Grade 11
  • Mathematics
  • A-SSE.B.3
  • Algebra
  • Free lesson

Aligned to A-SSE.B.3 — Common Core State Standards for Mathematics.

What this lesson teaches

Equivalent forms of an expression reveal different information. Completing the square rewrites a quadratic to expose its vertex (maximum or minimum).

Worked examples

Example 1

x² + 6x + 5 = (x+3)² − 4. The vertex form shows a minimum at (−3, −4).

Example 2

Rewrite 2x² − 12x + 7 by completing the square. Factor the 2 out of the x-terms first: 2(x² − 6x) + 7. Half of −6 is −3, and (−3)² = 9, so 2(x² − 6x + 9) − 2(9) + 7 = 2(x − 3)² − 11. The vertex is (3, −11) and the parabola opens upward.

Example 3

Common mistake: forgetting that the added square is multiplied by the leading coefficient. In 3(x² + 4x) + 1, adding 4 inside the parentheses adds 3 × 4 = 12 to the expression, so you must subtract 12, not 4: 3(x + 2)² − 12 + 1 = 3(x + 2)² − 11. Subtracting only 4 shifts the whole graph and changes the function.

Practice questions

  1. Complete the square for x² + 4x.
  2. What does vertex form tell you that standard form hides?
  3. Rewrite x² + 8x + 10 in vertex form.

Watch the lesson

Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.

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En español

Reescritura de Expresiones (Completar el Cuadrado)

Las formas equivalentes de una expresión revelan información diferente. Completar el cuadrado reescribe una expresión cuadrática para mostrar su vértice (máximo o mínimo).

Ejemplo 1: x² + 6x + 5 = (x+3)² − 4. La forma de vértice muestra un mínimo en (−3, −4).

Ejemplo 2: Reescribe 2x² − 12x + 7 completando el cuadrado. Primero factoriza el 2 de los términos con x: 2(x² − 6x) + 7. La mitad de −6 es −3, y (−3)² = 9, así que 2(x² − 6x + 9) − 2(9) + 7 = 2(x − 3)² − 11. El vértice es (3, −11) y la parábola abre hacia arriba.

Ejemplo 3: Error común: olvidar que el cuadrado añadido se multiplica por el coeficiente principal. En 3(x² + 4x) + 1, sumar 4 dentro del paréntesis agrega 3 × 4 = 12 a la expresión, así que hay que restar 12, no 4: 3(x + 2)² − 12 + 1 = 3(x + 2)² − 11. Restar solo 4 desplaza toda la gráfica y cambia la función.

More Algebra lessons

Zeros and Factors of PolynomialsA-APR.B.3Sum of a Finite Geometric SeriesA-SSE.B.4The Remainder TheoremA-APR.B.2Proving Polynomial IdentitiesA-APR.C.4Rewriting Rational ExpressionsA-APR.D.6Creating Equations to Model ProblemsA-CED.A.1

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