Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 9: solve (x + 3)(x − 8) = 0.
Answer
x = −3 or x = 8. Each bracket can be 0. - From this unit, Lesson 3: write x² + 6x + 1 as (x + 3)² + k. What is k?
Answer
−8. (x + 3)² = x² + 6x + 9, so k = 1 − 9 = −8. - From Extended Unit 1, Lesson 3: simplify √75.
Answer
5√3. 75 = 25 × 3.
Words used in this lesson
- discriminant
- b² − 4ac, the part under the root in the formula. For example: Δ = 49 − 40 = 9.
- real solution
- a solution that is an ordinary number. For example: x = 6 for x − 6 = 0.
- repeated solution
- one solution counted twice, when Δ = 0. For example: x = −2 for (x + 2)² = 0.
- perfect square
- a whole number that is a whole number squared. For example: 49 = 7².
- quadratic formula
- x = (−b ± √(b² − 4ac)) ÷ (2a). For example: solves ax² + bx + c = 0.
- irrational
- a number that cannot be written as a fraction. For example: √13.
The offer with a limit
A food stall in Marrakesh sells x kilograms of dates per day at a price that falls as x grows, so its daily income is x(20 − x) dirhams. The owner asks: “Is there any amount at which I make exactly 100 dirhams?” The question is whether an equation has a solution, before anyone solves it.
Predict first: for x² + bx + 9 = 0, as b changes, can the number of solutions change? If so, where? Then use the figure: the parabola on the left and the discriminant on the right.
x² + 9 = 0: Δ = 0² − 4(1)(9) = 0 − 36 = −36. Δ is negative, so there are no real solutions.
Show me the steps
- Set b = 4. Then Δ = 16 − 36 = −20, which is negative, so there are no real solutions. The lowest point of the parabola is 9 − 16 ÷ 4 = 5, above the x-axis.
- Set b = 6. Then Δ = 36 − 36 = 0, so there is one repeated solution, x = −3. The vertex sits exactly on the x-axis: the parabola touches it.
- Set b = 7. Then Δ = 49 − 36 = 13, which is positive but not a perfect square, so there are two irrational solutions, (−7 ± √13) ÷ 2: about −5.3 and −1.7.
- Set b = 10. Then Δ = 100 − 36 = 64 = 8², a perfect square, so the solutions are exact: −9 and −1. The equation factorizes as (x + 1)(x + 9) = 0.
The formula, derived. From ax² + bx + c = 0, divide by a and complete the square: (x + b ÷ (2a))² = (b² − 4ac) ÷ (4a²). Taking roots gives x = (−b ± √(b² − 4ac)) ÷ (2a).
The discriminant Δ = b² − 4ac sits under the root. Δ > 0: two real solutions. Δ = 0: one repeated solution (the vertex is on the axis). Δ < 0: no real solutions. If Δ is a perfect square, the solutions are rational and the quadratic factorizes with whole numbers.
Choose the method. Factorize when whole-number brackets are quick to see. Use the formula when they are not. Complete the square when the vertex is wanted too.
Make one side 0 first. (x − 1)(x + 5) = 7 does not mean x − 1 = 7 or x + 5 = 7. Expand, move the 7, and get x² + 4x − 12 = 0, which has solutions x = 2 and x = −6. Also, the ± in the formula gives two answers: do not drop one.
Worked examples
1. Solve 3x² − 5x − 2 = 0.
- ac = −6 and b = −5, so the pair is −6 and 1: 3x² − 6x + x − 2 = 3x(x − 2) + (x − 2).
- That is (3x + 1)(x − 2) = 0, so each bracket can be 0.
- x = −1/3 or x = 2.
2. Solve 2x² + 6x − 1 = 0, leaving the answer exact and then to two decimal places.
- Δ = 6² − 4(2)(−1) = 36 + 8 = 44, positive, so two solutions.
- x = (−6 ± √44) ÷ 4 = (−6 ± 2√11) ÷ 4 = (−3 ± √11) ÷ 2.
- With √11 ≈ 3.317: x ≈ 0.16 or x ≈ −3.16.
3. Solve x + 14 ÷ x = 9, where x is not 0.
- Multiply by x: x² + 14 = 9x, so x² − 9x + 14 = 0.
- (x − 2)(x − 7) = 0, so x = 2 or x = 7.
- Check x = 2: 2 + 7 = 9. Check x = 7: 7 + 2 = 9. Both work.
Your go. The first step is given. For which values of k has x² + 10x + k = 0 no real solutions?
- Δ = 10² − 4(1)(k) = 100 − 4k.
- No real solutions needs Δ < 0. Solve that inequality for k.
Check the answer
100 − 4k < 0 gives 4k > 100, so k > 25. At k = 25 the parabola touches the axis at x = −5, and above that it floats clear of it.
With a GDC (optional)
A GDC can check a solution. Graph y = 2x² + 6x − 1 and use the zero tool (under the graph or calculate menu) to find where it meets the x-axis: it gives about 0.16 and −3.16. Many GDCs also have a polynomial solver in the equation menu.
A GDC gives decimals, and the question may ask for an exact form such as (−3 ± √11) ÷ 2, so keep the working from the formula. Everything here can be done by hand.
Check before you go on
1. 3x² + 5x + k = 0 has no real solutions when:
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. (x − 2)(x + 5) = 18. What is the positive solution?
3. The formula for 2x² − 8x + 5 = 0 gives x = (8 ± √D) ÷ 4. Find D.
Level 2 · Building
4. Solve 2x² − 7x − 15 = 0. What is the positive solution?
5. x² + bx + 16 = 0 has exactly one solution. What is the positive value of b?
Level 3 · Stretch
6. kx² + 6x + 3 = 0, where k ≠ 0, has two different real solutions exactly when k is less than a certain number. What is that number?
7. x + 12 ÷ x = 7, where x is not 0. What is the smaller solution?
Something to think about
In x² + bx + 9, the equation has a repeated solution at two values of b. What are they, and what do the two parabolas look like? Then say what the repeated solution has to do with the vertex.
One way to think about it
The repeated solution occurs where Δ = b² − 36 = 0, so b = 6 and b = −6. The parabolas are mirror images in the y-axis, and each touches the x-axis at its vertex (x = −3 for b = 6 and x = 3 for b = −6): a repeated solution is the vertex lying exactly on the axis, where the two zeros have merged into one.
Exit ticket
8. Which equation has exactly one real solution?