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Completing the square for quadratic expressions

How to complete the square for a quadratic expression such as \(a \cdot x^2 + b \cdot x + c\)?

Subject: Math Algebra Chapter: Equations Topic: Quadratic Equation Solver Answer included
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Accepted answer Answer included

How to complete the square is a rewrite of a quadratic into a perfect-square form. The standard algebraic target is a vertex form \(a \cdot (x + h)^2 + k\), which is equivalent to \(a \cdot x^2 + b \cdot x + c\) and reveals the parabola’s vertex and many solution methods immediately.

Perfect-square identity

The central pattern is the expansion of a binomial square:

\[ (x + p)^2 = x^2 + 2 \cdot p \cdot x + p^2 . \]

A quadratic expression contains a perfect-square trinomial whenever the constant term matches the square of half the coefficient on the linear term (after normalization).

Monic quadratic \(x^2 + b \cdot x + c\)

For a leading coefficient \(1\), the linear coefficient \(b\) fixes the shift \(p\) by matching \(2 \cdot p = b\), so \(p = \frac{b}{2}\). The rewrite introduces \(\left(\frac{b}{2}\right)^2\) and compensates with subtraction to preserve equality:

\[ x^2 + b \cdot x + c = \left(x + \frac{b}{2}\right)^2 + \left(c - \left(\frac{b}{2}\right)^2\right). \]

The constant adjustment is \(c - \left(\frac{b}{2}\right)^2\). No numerical value is “added” without also being “subtracted”; the expression is reorganized into a square plus a remainder.

General quadratic \(a \cdot x^2 + b \cdot x + c\)

When \(a \neq 1\), normalization places a monic quadratic inside parentheses. Factoring \(a\) from the quadratic terms yields:

\[ a \cdot x^2 + b \cdot x + c = a \cdot \left(x^2 + \frac{b}{a} \cdot x\right) + c . \]

The perfect square is formed inside the parentheses using \(\frac{b}{a}\) as the linear coefficient:

\[ x^2 + \frac{b}{a} \cdot x = \left(x + \frac{b}{2 \cdot a}\right)^2 - \left(\frac{b}{2 \cdot a}\right)^2 . \]

Substituting back gives the completed-square (vertex) form:

\[ a \cdot x^2 + b \cdot x + c = a \cdot \left(x + \frac{b}{2 \cdot a}\right)^2 + \left(c - \frac{b^2}{4 \cdot a}\right). \]
Standard form Completed-square form Vertex parameters
\(x^2 + b \cdot x + c\) \(\left(x + \frac{b}{2}\right)^2 + \left(c - \left(\frac{b}{2}\right)^2\right)\) \(h = \frac{b}{2}\), \(k = c - \left(\frac{b}{2}\right)^2\)
\(a \cdot x^2 + b \cdot x + c\) \(a \cdot \left(x + \frac{b}{2 \cdot a}\right)^2 + \left(c - \frac{b^2}{4 \cdot a}\right)\) \(h = \frac{b}{2 \cdot a}\), \(k = c - \frac{b^2}{4 \cdot a}\)

Worked example

Consider \(x^2 + 6 \cdot x + 5\). The half-linear coefficient is \(\frac{6}{2} = 3\), so the square term is \(3^2 = 9\):

\[ x^2 + 6 \cdot x + 5 = \left(x + 3\right)^2 + \left(5 - 9\right) = \left(x + 3\right)^2 - 4 . \]
Completing the square as a geometric area model A square of side x is extended by two rectangles of area 3x each, leaving a missing corner of area 9; adding the 3-by-3 corner completes a larger square of side x+3. 3x 3x missing 9 side x + 3 x + 3 x² + 6x needs +9 to become (x + 3)² complete 3x 3x 9 side x + 3 x + 3 (x + 3)² = x² + 6x + 9
The area model shows why \(\left(\frac{b}{2}\right)^2\) is the needed addition for the monic case \(x^2 + b \cdot x\). For \(b = 6\), the missing corner is \(9\), giving \((x + 3)^2\). Any extra constant \(c\) remains as a vertical shift after the square is formed.

Solving a quadratic after square completion

The completed-square form reduces many equations to a single square-root operation. For example, the equation \(x^2 + 6 \cdot x + 5 = 0\) becomes:

\[ (x + 3)^2 - 4 = 0 \quad \Longleftrightarrow \quad (x + 3)^2 = 4 \quad \Longleftrightarrow \quad x + 3 = \pm 2, \]
\[ x = -3 \pm 2 \quad \Longleftrightarrow \quad x = -1 \ \text{or}\ x = -5 . \]

Common pitfalls

  • Half-coefficient mismatch: the linear term in \((x + p)^2\) is \(2 \cdot p \cdot x\), so \(p\) equals one-half of the linear coefficient after normalization.
  • Leading coefficient handling: for \(a \cdot x^2 + b \cdot x + c\), the square must be formed inside the factor \(a \cdot (\cdot)\); the adjustment outside becomes \(c - \frac{b^2}{4 \cdot a}\).
  • Sign control: the shift in \((x - h)^2\) corresponds to \(-2 \cdot h \cdot x\); a positive linear term corresponds to \((x + h)^2\) in the monic case.

Summary form

The canonical algebraic result used for how to complete the square is:

\[ a \cdot x^2 + b \cdot x + c = a \cdot \left(x + \frac{b}{2 \cdot a}\right)^2 + \left(c - \frac{b^2}{4 \cdot a}\right), \quad a \neq 0. \]
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