1.1.1 Completing the Square

In this topic we will learn how to:

  • carry out the process of completing the square for a quadratic polynomial ax2+bx+cax^{2} + bx + c and use a completed square form

There are two main scenarios, when completing the square. The first scenario is when the coefficient of x2x^{2} is 11 and the second scenario is when the coefficient of x2x^{2} is not 11. Let’s see how we go about the two scenarios:


Completing the square when the coefficient of x2 is 1\textcolor{gray}{\textbf{Completing the square when the}\\ \textbf{ coefficient of }x^{2}\textbf{ is }1}The general form for completing the square when the coefficient of x2x^{2} is 11, where the quadratic is x2+bx+cx^{2} + bx + c :
(x+b2)2(b2)2+c\left(x + \frac{b}{2}\right)^{2} - \left(\frac{b}{2}\right)^{2} + cLet’s look at an example.

Example 1
Complete the square for x2+5x6x^{2} + 5x - 6.

The first step is to rearrange the equation into the form x2+bx+cx^{2} + bx + c,
x2+5x 6x^{2} + \textcolor{#2192ff}{5}x \textcolor{#0f0}{-\ 6}Identify bb and cc,
b=5,  c=6\textcolor{#2192ff}{b} = \textcolor{#2192ff}{5}, \ \ \textcolor{#0f0}{c} = \textcolor{#0f0}{-6}
Divide b\textcolor{#2192ff}{b} by 22,
52\textcolor{#2192ff}{\frac{5}{2}}Substitute into the general completed square form,
(x+52)2(52)2+(6)\left(x + \textcolor{#2192ff}{\frac{5}{2}}\right)^{2} - \left(\textcolor{#2192ff}{\frac{5}{2}}\right)^{2} + \textcolor{#0f0}{(- 6)}Simplify,
(x+52)2494\left(x + \frac{5}{2}\right)^{2} -\frac{49}{4}


Completing the square when the coefficient of x2 is not 1\textcolor{gray}{\textbf{Completing the square when the}\\ \textbf{ coefficient of }x^{2}\textbf{ is not 1}}Let’s look at an example.

Example 2
Complete the square for 2x2+9x+102x^{2} + 9x + 10.

The first step is to rearrange the equation into the form ax2+bx+cax^{2} + bx + c,
2x2+9x+102x^{2} + 9x + 10Factor out the coefficient of x2x^{2} from the first two terms,
2(x2+92x)+102\textcolor{#2192ff}{\left(x^{2} + \frac{9}{2}x\right)} + 10Complete the square for the quadratic inside parentheses using the method outlined above i.e Completing the square when the coefficient of x2x^{2} is 11,
2[(x+94)2(94)2]+102\textcolor{#2192ff}{\left[\left(x + \frac{9}{4}\right)^{2}- \left(\frac{9}{4}\right)^{2}\right]} + 10Simplify inside the square brackets,
2[(x+94)28116]+102\textcolor{#2192ff}{\left[\left(x + \frac{9}{4}\right)^{2}- \frac{81}{16}\right]} + 10Remove the square brackets,
2(x+94)2818+102\left(x + \frac{9}{4}\right)^{2} - \frac{81}{8} + 10Simplify,
2(x+94)2182\left(x + \frac{9}{4}\right)^{2} - \frac{1}{8}


Locating the vertex from the completed square form\textcolor{gray}{\textbf{Locating the vertex from the}\\ \textbf{ completed square form}}We can also use the completed square form to determine the coordinates of the vertex.

Given the completed square form, (x+b)2+c(x + b)^{2} + c, the vertex is,
(b,c)(-b, c)Note: The vertex is also known as the turning point, stationary point, maximum or minimum point.


Let’s look at some past paper questions on completing the square

1. Functions ff and gg are both defined for xRx \in \mathbb{R} and are given by
f(x)=x24x+9f(x) = x^{2} - 4x + 9g(x)=2x2+4x+12g(x) = 2x^{2} + 4x + 12Express f(x)f(x) in the form (xa)2+b(x - a)^{2} + b. (9709/11/O/N/22 number 9)
f(x)=x24x+9f(x) = x^{2} - 4x + 9The first step is to arrange the equation in the form ax2+bx+cax^{2} + bx + c,
x2 4x+9x^{2} \textcolor{#2192ff}{-\ 4}x + \textcolor{#0f0}{9}Identify bb and cc,
b=4,  c=9\textcolor{#2192ff}{b} = \textcolor{#2192ff}{-4}, \ \ \textcolor{#0f0}{c} = \textcolor{#0f0}{9}Recall the formula for completing the square,
(x+b2)2(b2)2+c\left(x + \frac{\textcolor{#2192ff}{b}}{2}\right)^{2} - \left(\frac{\textcolor{#2192ff}{b}}{2}\right)^{2} + \textcolor{#0f0}{c}Substitute into the formula,
(x+42)2(42)2+9\left(x + \frac{\textcolor{#2192ff}{-4}}{2}\right)^{2} - \left(\frac{\textcolor{#2192ff}{-4}}{2}\right)^{2} + \textcolor{#0f0}{9}Simplify,
(x2)24+9\left(x - 2\right)^{2} - 4 + 9(x2)2+5\left(x - 2\right)^{2} + 5Therefore the final answer is,
f(x)=(x2)2+5f(x) = \left(x - 2\right)^{2} + 5

2. Express x28x+11x^{2} - 8x + 11 in the form (x+p)2+q(x + p)^{2} + q where pp and qq are constants. (9709/11/M/J/22 number 1)x28x+11x^{2} - 8x + 11The first step is to arrange the equation in the form ax2+bx+cax^{2} + bx + c,
x2 8x+11x^{2} \textcolor{#2192ff}{-\ 8}x + \textcolor{#0f0}{11}Identify bb and cc,
b=8,  c=11\textcolor{#2192ff}{b} = \textcolor{#2192ff}{-8}, \ \ \textcolor{#0f0}{c} = \textcolor{#0f0}{11}Recall the formula for completing the square,
(x+b2)2(b2)2+c\left(x + \frac{\textcolor{#2192ff}{b}}{2}\right)^{2} - \left(\frac{\textcolor{#2192ff}{b}}{2}\right)^{2} + \textcolor{#0f0}{c}Substitute into the formula,
(x+82)2(82)2+11\left(x + \frac{\textcolor{#2192ff}{-8}}{2}\right)^{2} - \left(\frac{\textcolor{#2192ff}{-8}}{2}\right)^{2} + \textcolor{#0f0}{11}Simplify,
(x4)216+11\left(x - 4\right)^{2} - 16 + 11(x4)25\left(x - 4\right)^{2} - 5Therefore the final answer is,
f(x)=(x4)25f(x) = \left(x - 4\right)^{2} - 5

3. The function ff is defined by f(x)=2x216x+23f(x) = 2x^{2} - 16x + 23 for x<3x < 3. Express f(x)f(x) in the form 2(x+a)2+b2(x + a)^{2} + b. (9709/13/M/J/22 number 6)f(x)=2x216x+23f(x) = 2x^{2} - 16x + 23The first step is to arrange the equation into the form ax2+bx+cax^{2} + bx + c,
2x216x+232x^{2} - 16x + 23Factor out the coefficient of x2x^{2} from the first two terms,
2(x28x)+232\textcolor{#2192ff}{\left(x^{2} - 8x\right)} + 23Complete the square for the quadratic inside parentheses,
2[(x4)2(4)2]+232\textcolor{#2192ff}{\left[\left(x - 4\right)^{2}- \left(4\right)^{2}\right]} + 23Simplify inside the square brackets,
2[(x4)216]+232\textcolor{#2192ff}{\left[\left(x - 4\right)^{2}- 16\right]} + 23Remove the square brackets,
2(x4)232+232\left(x - 4\right)^{2} - 32 + 23Simplify,
2(x4)292\left(x - 4\right)^{2} - 9Therefore the final answer is,
2(x4)292\left(x - 4\right)^{2} - 9

4. The equation of a curve is y=4x2+20x+6y = 4x^{2} + 20x + 6. Express the equation in the form y=a(x+b)2+cy = a(x + b)^{2} + c, where aa, bb, and cc are constants. (9709/12/O/N/22 number 6)y=4x2+20x+6y = 4x^{2} + 20x + 6The first step is to arrange the equation into the form ax2+bx+cax^{2} + bx + c,
4x2+20x+64x^{2} + 20x + 6Factor out the coefficient of x2x^{2} from the first two terms,
4(x2+5x)+64\textcolor{#2192ff}{\left(x^{2} + 5x\right)} + 6Complete the square for the quadratic inside parentheses,
4[(x+52)2(52)2]+64\textcolor{#2192ff}{\left[\left(x + \frac{5}{2}\right)^{2}- \left(\frac{5}{2}\right)^{2}\right]} + 6Simplify inside the square brackets,
4[(x+52)2254]+64\textcolor{#2192ff}{\left[\left(x + \frac{5}{2}\right)^{2}- \frac{25}{4}\right]} + 6Remove the square brackets,
4(x+52)225+64\left(x + \frac{5}{2}\right)^{2} - 25 + 6Simplify,
4(x+52)2194\left(x + \frac{5}{2}\right)^{2} - 19Therefore the final answer is,
4(x+52)2194\left(x + \frac{5}{2}\right)^{2} - 19