Completing the square Knowledge-Based Study

Study notes

1. What is Completing the Square?

   Completing the square is an algebraic technique used to rewrite a quadratic expression of the form $ax^2 + bx + c$ into vertex form, $a(x - h)^2 + k$, or to solve quadratic equations.

   The method is based on the algebraic identity for a perfect square trinomial:

   $$(x + d)^2 = x^2 + 2dx + d^2$$$$(x - d)^2 = x^2 - 2dx + d^2$$

   Notice that in any monic perfect square trinomial (where the coefficient of $x^2$ is 1), the constant term is always the square of half the coefficient of $x$:

   $$\left(\frac{2d}{2}\right)^2 = d^2$$

   When given an incomplete quadratic expression such as $x^2 + bx$, we can complete the square by adding $\left(\frac{b}{2}\right)^2$.

   Step-by-Step Procedure for $a = 1$

   To convert an expression of the form $x^2 + bx + c$ into completed square form:

   1. Identify the linear coefficient, $b$.
   2. Divide $b$ by 2 and square the result: $\left(\frac{b}{2}\right)^2$.
   3. Add and subtract this value within the expression to keep the value unchanged: $$x^2 + bx + c = \left(x^2 + bx + \left(\frac{b}{2}\right)^2\right) - \left(\frac{b}{2}\right)^2 + c$$
   4. Factor the perfect square trinomial and simplify the remaining constants: $$\left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$$

   Worked Example

   Rewrite $x^2 - 6x + 4$ in completed square form.

   • Here, $b = -6$.
   • Half of $b$ is $\frac{-6}{2} = -3$, and its square is $(-3)^2 = 9$.
   • Add and subtract 9: $$x^2 - 6x + 9 - 9 + 4$$
   • Factor the first three terms: $$(x - 3)^2 - 5$$

   Handling Leading Coefficients Other Than 1

   When the leading coefficient $a \neq 1$, you must factor $a$ out of the quadratic and linear terms before finding the term needed to complete the square.

   1. Group the quadratic and linear terms: $$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x\right) + c$$
   2. Find half of the new linear coefficient, $\frac{b}{2a}$, and square it: $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$.
   3. Add and subtract this term inside the parentheses: $$a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c$$
   4. Distribute $a$ when moving the subtracted term outside: $$a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b^2}{4a^2}\right) + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$

   Worked Example

   Rewrite $2x^2 + 12x + 7$ by completing the square.

   • Factor 2 from the first two terms: $$2(x^2 + 6x) + 7$$
   • Half of 6 is 3, and $3^2 = 9$. Add and subtract 9 inside the bracket: $$2(x^2 + 6x + 9 - 9) + 7$$
   • Expand the bracket for $-9$: $$2(x + 3)^2 - 2(9) + 7 = 2(x + 3)^2 - 18 + 7 = 2(x + 3)^2 - 11$$

   1. Solving Quadratic Equations

   To solve an equation such as $x^2 + 8x - 9 = 0$ by completing the square:

   1. Isolate the constant term on the right-hand side: $$x^2 + 8x = 9$$
   2. Add $\left(\frac{8}{2}\right)^2 = 16$ to both sides: $$x^2 + 8x + 16 = 9 + 16$$
   3. Factor the left side: $$(x + 4)^2 = 25$$
   4. Take the square root of both sides: $$x + 4 = \pm 5$$
   5. Solve for $x$: $$x = -4 + 5 = 1 \quad \text{or} \quad x = -4 - 5 = -9$$

   2. Locating the Vertex and Extreme Value

   Writing a quadratic function in vertex form $y = a(x - h)^2 + k$ directly reveals the coordinates of the turning point (vertex) at $(h, k)$:

   • If $a > 0$, the parabola opens upwards and has a minimum value of $k$ at $x = h$.
   • If $a < 0$, the parabola opens downwards and has a maximum value of $k$ at $x = h$.

   For example, the function $f(x) = (x - 3)^2 - 5$ has a minimum value of $-5$ occurring at $x = 3$, giving vertex coordinates $(3, -5)$.