Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 9: expand (x + 5)².
Answer
x² + 10x + 25. The middle term is 2 × 5x and the last is 5². - From Year 10, Unit 1: simplify √50.
Answer
5√2, because 50 = 25 × 2. - Work out √(64 + 36).
Answer
10. Add under the root first: √100 = 10.
Words used in this lesson
- completing the square
- adding the missing corner so an expression becomes a perfect square. For example: turning x² + 2x into (x + 1)² − 1.
- perfect square
- an expression that is something squared. For example: (x + 9)².
- quadratic formula
- a formula that gives the solutions of ax² + bx + c = 0. For example: uses a, b and c.
- exact answer
- an answer that keeps surds and fractions. For example: 2 + √3.
- coefficient
- the number in front of a letter. For example: the 6 in 6x².
- plus or minus
- the symbol ± stands for two values at once. For example: 5 ± 2 is 3 and 7.
When the pairs run out
A groundsman in Nairobi marks out a seed bed that is 4 m longer than it is wide, with an area of 3 m². Its width x metres satisfies x(x + 4) = 3, which is x² + 4x − 3 = 0. Look for two whole numbers with product −3 and sum 4: 1 and −3 give −2, and −1 and 3 give 2. There is no pair.
Predict first: does that mean the equation has no solutions, or only that factorizing is the wrong tool? What could you do instead? Then use the figure to see how a square can be completed.
x² + 6x + 5 = 0: completing the square gives (x + 3)² = 4. So x = −3 ± 2, which is x = −5 or x = −1.
Show me the steps
- Set m = 4 and c = 7. x² + 8x + 7 = 0 completes to (x + 4)² = 9, so x = −4 ± 3, which is −7 or −1.
- Set m = 2 and c = 1. (x + 2)² = 3, so x = −2 ± √3. These are not whole numbers: about −3.73 and −0.27.
- Set m = 3 and c = 9. (x + 3)² = 0, so there is one solution, x = −3, and the graph just touches the x-axis.
- Set m = 1 and c = 5. (x + 1)² = −4, and a square cannot be negative, so there is no real solution.
Completing the square. x² + bx is a square of side x plus two strips each b ÷ 2 wide. It becomes a perfect square when the missing corner (b ÷ 2)² is added: x² + bx + (b ÷ 2)² = (x + b ÷ 2)². Add the same amount to both sides of the equation, take the square root of both sides, and remember the ±.
The quadratic formula. For ax² + bx + c = 0, doing this once with letters gives x = (−b ± √(b² − 4ac)) ÷ (2a). Put the brackets in: the whole of the top is divided by 2a.
Leave an answer exact (with a surd) unless the question asks for decimal places. Both methods give the same solutions.
Two common mistakes. Forgetting the ± when taking a square root: (x + 4)² = 19 has two solutions, −4 + √19 and −4 − √19. And writing −b ÷ 2a without brackets, or taking the 2a only under the root: the 2a divides the whole top line.
Worked examples
1. Solve x² + 8x − 3 = 0 by completing the square.
- Move the constant: x² + 8x = 3. Half of 8 is 4, and 4² = 16, so add 16 to both sides.
- (x + 4)² = 3 + 16 = 19.
- x + 4 = ±√19, so x = −4 ± √19, which is about 0.36 or −8.36.
2. Solve 3x² − 7x + 1 = 0 with the formula, to 2 decimal places.
- a = 3, b = −7, c = 1. Then b² − 4ac = 49 − 12 = 37.
- x = (7 ± √37) ÷ 6, because −b = 7 and 2a = 6.
- √37 = 6.083, so x = 13.083 ÷ 6 = 2.18 or x = 0.917 ÷ 6 = 0.15.
Your go. The first step is given. Solve x² − 5x + 2 = 0 with the formula, to 2 decimal places.
- a = 1, b = −5, c = 2. Write b² − 4ac with brackets: (−5)² − 4(1)(2).
- Work it out, then put it in the formula with −b = 5 and 2a = 2.
Check the answer
25 − 8 = 17, so x = (5 ± √17) ÷ 2. With √17 = 4.123, the solutions are 4.56 and 0.44.
With a GDC (optional)
A GDC can solve a quadratic equation in decimals. Open the polynomial or equation solver, enter the coefficients a, b and c, and read both roots. For x² − 12x + 7 = 0 it shows about 11.39 and 0.61, which agree with the exact answer 6 ± √29.
The solver gives decimals, so it does not show the exact form. Give the exact form from your hand working, and use the GDC to check it and to round. Type the whole formula with brackets, such as (12 + √(144 − 28)) ÷ 2. Everything here can be done by hand.
Check before you go on
1. Which is x² − 10x + 3 written in the form (x − p)² + q?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. Completing the square on x² + 14x − 15 = 0 gives (x + 7)² = k. Find k.
3. Use the quadratic formula to solve 2x² + 3x − 4 = 0. Give the larger solution to 2 decimal places.
Level 2 · Building
4. Solve x² − 4x − 6 = 0 by completing the square. Give the positive solution to 2 decimal places.
5. A student in Manila launches a water rocket straight up from a platform 1.5 m high at 20 m/s. Taking gravity as 10 m/s², its height after t seconds is h = −5t² + 20t + 1.5. How many seconds until it lands (h = 0)? Give the positive answer to 2 decimal places.
Level 3 · Stretch
6. x² − 8x + c = 0 has the solution x = 4 + √5. Find c.
7. 2x² + 12x + 5 can be written as 2(x + 3)² + q. Find q.
Something to think about
Complete the square on x² + bx + c = 0 for general b and c. What does the expression under the square root become, and when is it negative? Try b = 6 and b = 2 with c = 5.
One way to think about it
Move c across and add (b ÷ 2)² to both sides: (x + b ÷ 2)² = (b ÷ 2)² − c, which is (b² − 4c) ÷ 4. So x = −b ÷ 2 ± √(b² − 4c) ÷ 2. For b = 6 and c = 5 the root part is √16 ÷ 2 = 2 and the solutions are −5 and −1. For b = 2 and c = 5 it is √(−16), which does not exist, so there are no real solutions. The expression under the root, b² − 4c, decides everything; Lesson 6 gives it a name.
Exit ticket
8. Which gives the exact solutions of x² − 6x + 2 = 0?