In this topic we will learn how to:
- carry out the process of completing the square for a quadratic polynomial ax2+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 x2 is 1 and the second scenario is when the coefficient of x2 is not 1. Let’s see how we go about the two scenarios:
Completing the square when the coefficient of x2 is 1The general form for completing the square when the coefficient of x2 is 1, where the quadratic is x2+bx+c :
(x+2b)2−(2b)2+cLet’s look at an example.
Example 1
Complete the square for x2+5x−6.
The first step is to rearrange the equation into the form x2+bx+c,
x2+5x− 6Identify b and c,
b=5, c=−6
Divide b by 2,
25Substitute into the general completed square form,
(x+25)2−(25)2+(−6)Simplify,
(x+25)2−449
Completing the square when the coefficient of x2 is not 1Let’s look at an example.
Example 2
Complete the square for 2x2+9x+10.
The first step is to rearrange the equation into the form ax2+bx+c,
2x2+9x+10Factor out the coefficient of x2 from the first two terms,
2(x2+29x)+10Complete the square for the quadratic inside parentheses using the method outlined above i.e Completing the square when the coefficient of x2 is 1,
2[(x+49)2−(49)2]+10Simplify inside the square brackets,
2[(x+49)2−1681]+10Remove the square brackets,
2(x+49)2−881+10Simplify,
2(x+49)2−81
Locating the vertex from the completed square formWe can also use the completed square form to determine the coordinates of the vertex.
Given the completed square form, (x+b)2+c, the vertex is,
(−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 f and g are both defined for x∈R and are given by
f(x)=x2−4x+9g(x)=2x2+4x+12Express f(x) in the form (x−a)2+b. (9709/11/O/N/22 number 9)f(x)=x2−4x+9The first step is to arrange the equation in the form ax2+bx+c,
x2− 4x+9Identify b and c,
b=−4, c=9Recall the formula for completing the square,
(x+2b)2−(2b)2+cSubstitute into the formula,
(x+2−4)2−(2−4)2+9Simplify,
(x−2)2−4+9(x−2)2+5Therefore the final answer is,
f(x)=(x−2)2+5
2. Express x2−8x+11 in the form (x+p)2+q where p and q are constants. (9709/11/M/J/22 number 1)x2−8x+11The first step is to arrange the equation in the form ax2+bx+c,
x2− 8x+11Identify b and c,
b=−8, c=11Recall the formula for completing the square,
(x+2b)2−(2b)2+cSubstitute into the formula,
(x+2−8)2−(2−8)2+11Simplify,
(x−4)2−16+11(x−4)2−5Therefore the final answer is,
f(x)=(x−4)2−5
3. The function f is defined by f(x)=2x2−16x+23 for x<3. Express f(x) in the form 2(x+a)2+b. (9709/13/M/J/22 number 6)f(x)=2x2−16x+23The first step is to arrange the equation into the form ax2+bx+c,
2x2−16x+23Factor out the coefficient of x2 from the first two terms,
2(x2−8x)+23Complete the square for the quadratic inside parentheses,
2[(x−4)2−(4)2]+23Simplify inside the square brackets,
2[(x−4)2−16]+23Remove the square brackets,
2(x−4)2−32+23Simplify,
2(x−4)2−9Therefore the final answer is,
2(x−4)2−9
4. The equation of a curve is y=4x2+20x+6. Express the equation in the form y=a(x+b)2+c, where a, b, and c are constants. (9709/12/O/N/22 number 6)y=4x2+20x+6The first step is to arrange the equation into the form ax2+bx+c,
4x2+20x+6Factor out the coefficient of x2 from the first two terms,
4(x2+5x)+6Complete the square for the quadratic inside parentheses,
4[(x+25)2−(25)2]+6Simplify inside the square brackets,
4[(x+25)2−425]+6Remove the square brackets,
4(x+25)2−25+6Simplify,
4(x+25)2−19Therefore the final answer is,
4(x+25)2−19