A quadratic such as 2x2−8x+3 hides its most useful facts. Where is its lowest point? What is its smallest value? Completing the square rewrites it in a form where those answers are sitting in plain view. It is one of the most used skills in Pure 1: you will need it for ranges of functions, for circles, for inverse functions and for maximum and minimum problems.
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Solve by completing the square | Step by Step Technique
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Completing the Square in A Level Maths - Step-by-Step Worked Example
3If a=1, take a out of the x2 and x terms first, then complete the square inside the bracket.
4The vertex of y=a(x+p)2+q is (−p,q) and the line of symmetry is x=−p.
5If a>0 the vertex is a minimum point; if a<0 it is a maximum point.
x
+
3)2=
x2+
6x+
9
So x2+6x is "almost" a perfect square. It is (x+3)2 with the extra 9 taken away:
x2+6x=(x+3)2−9
The rule: halve the coefficient of x, put it in the bracket, then subtract its square.
x2+bx=(x+2b)2−(2b)2
The finished form a(x+p)2+q is called completed square form. Here is why it is so useful. A square, (x+p)2, can never be negative, and it equals zero only when x=−p. So:
If a>0, the smallest value of a(x+p)2+q is q, reached at x=−p.
If a<0, the largest value is q, reached at x=−p.
The point (−p,q) is the vertex (the turning point) of the parabola, and the vertical line x=−p is its line of symmetry.
When a=1, factor a out of the x2 and x terms only. Leave the constant outside.
Worked example
Example 1. Express 2x2−8x+3 in the form a(x+p)2+q. Hence state the coordinates of the vertex of y=2x2−8x+3 and sketch the curve.
Take out the factor 2 from the first two terms:
2x2−8x+3=2(x2−4x)+3
Complete the square inside the bracket. Half of −4 is −2:
x2−4x=(x−2)2−4
Substitute back and tidy up. Remember the 2 multiplies the −4 too:
2[(x−2)2−4]+3=2(x−2)2−8+3=2(x−2)2−5
So a=2, p=−2, q=−5. Since a>0 the curve is U-shaped with a minimum at the vertex (2,−5). The line of symmetry is x=2. Putting x=0 gives the y-intercept, 3. Setting y=0:
2(x−2)2=5⇒x−2=±25⇒x=2±210
That is x≈0.42 and x≈3.58.
Graph of y = 2(x − 2)² − 5: a U-shaped parabola with minimum point (2, −5), line of symmetry x = 2, y-intercept 3 and roots at about x = 0.42 and x = 3.58
Example 2. Find the greatest value of 5+6x−3x2 and the value of x where it occurs.
Write the x2 term first and take out −3:
−3x2+6x+5=−3(x2−2x)+5
Complete the square: x2−2x=(x−1)2−1. Then
−3[(x−1)2−1]+5=−3(x−1)2+3+5=8−3(x−1)2
The term 3(x−1)2 is never negative, so the expression is largest when that term is zero. The greatest value is 8, when x=1.
Common mistakes
Forgetting to multiply out. In 2[(x−2)2−4] the 2 multiplies the −4 as well, giving −8, not −4.
Sign of the vertex.(x−2)2 gives a vertex at x=+2, not −2. Ask "what makes the bracket zero?"
Factoring a out of the constant. Write 2(x2−4x)+3. Dividing the 3 by 2 as well is allowed but creates fractions and errors.
Max or min? Check the sign of a before calling the vertex a minimum.
Where you see this in real life
Anything that follows a parabola has a best or worst point: the top of a ball's flight, the price that gives a company the most profit, or the speed at which a car uses the least fuel. Completing the square finds that point exactly, without needing calculus. The same idea also turns the messy equation of a circle into one that shows its centre and radius.