Completing the square: factorising and solving equations
Mathematics Prep 2 Term 1, Unit 2, Lesson 4
GoalComplete the square by adding the missing term, and use it to factorise and to solve equations whose roots are not whole numbers.
New words
completing the square · adding the term that turns a trinomial into a perfect squarehalf the coefficient · take half of the x-coefficient and square itbalance · what you add you must also subtract (or add to both sides)no real root · a square equal to a negative number has no solution in R
Explanation
x² + 10x is not yet a perfect square. The picture shows a square x by x and two strips 5 by x; the missing corner is 5 × 5 = 25, which is (half of 10)². Adding it gives x² + 10x + 25 = (x + 5)². To use this in an equation, add the same number to both sides. Solve x² − 8x + 3 = 0: move the 3, x² − 8x = −3; add (8/2)² = 16 to both sides, x² − 8x + 16 = 13; so (x − 4)² = 13 and x − 4 = ±√13, which gives x = 4 + √13 or x = 4 − √13. These roots are not whole numbers, but they are real. If you get (x + 1)² = −4, there is no real solution, because a square is never negative, and the solution set is ∅. Completing the square can also factorise: x⁴ + 4y⁴ = (x² + 2y²)² − (2xy)² = (x² + 2y² − 2xy)(x² + 2y² + 2xy).
Worked examples
Solve x² + 6x + 2 = 0. Move 2: x² + 6x = −2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 7. So (x + 3)² = 7, x + 3 = ±√7, and x = −3 + √7 or x = −3 − √7.
A rectangle is 4 m longer than it is wide and its area is 12 m². With width w: w² + 4w = 12. Add 4: w² + 4w + 4 = 16, so (w + 2)² = 16 and w + 2 = 4 (a width cannot be negative). So w = 2 m and the length is 6 m. Check: 2 × 6 = 12.
Solve x² + 2x + 5 = 0. Then x² + 2x + 1 = −4, so (x + 1)² = −4. A square cannot be negative, so there is no real solution: the solution set is ∅.
Class activity
Square corners. With squared paper, draw a square x by x with x = 6 squares and two strips of 3 by 6. How many small squares make the missing corner? What does this tell you about x² + 6x? Repeat for x² + 8x with x = 8 squares.
Practice
1
What must be added to x² + 12x to make a perfect square?
(12/2)² = 36, giving (x + 6)².
2
Solve x² − 6x + 4 = 0.
x² − 6x + 9 = 5, (x − 3)² = 5, so x = 3 ± √5.
3
Solve x² + 4x − 1 = 0.
x² + 4x + 4 = 5, (x + 2)² = 5, so x = −2 ± √5.
4
Solve x² − 2x + 7 = 0 in R.
(x − 1)² = −6 has no real solution, so the set is ∅.
5
Hana solves x² + 6x = 7 by writing (x + 3)² = 7. What is wrong?
She added 9 on the left only. Add 9 to both sides: (x + 3)² = 16, so x + 3 = ±4 and x = 1 or x = −7.
