Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 10, Lesson 5: how many solutions does (x + 2)² = 9 have, and what are they?
Answer
Two: x = 1 and x = −5. Take the square root of both sides, with ±3. - From Year 8: work out (−6)² and then −6².
Answer
36 and −36. The brackets decide whether the minus sign is squared. - Work out 20 − 4 × 2 × (−3).
Answer
44. −4 × 2 × (−3) is +24, and 20 + 24 = 44.
Words used in this lesson
- discriminant
- the value b² − 4ac that tells how many real solutions a quadratic has. For example: Δ = 9 − 16 for x² + 3x + 4.
- real solution
- a solution that is an ordinary number on the number line. For example: x = 2.5.
- repeated solution
- a solution counted twice because the factor is squared. For example: x = 5 in (x − 5)² = 0.
- perfect square
- the square of a whole number. For example: 121 is 11².
- rational
- a number that can be written as a fraction of integers. For example: 0.75 = 3/4.
- parameter
- a letter in a formula whose value can be chosen. For example: k in x² + kx + 1.
Before you solve
A quadratic model of a fountain jet, a bridge cable or a profit curve may or may not reach a given level. Solving takes time, and sometimes the answer is that there is no solution at all. Is there a quick test that tells you how many real solutions to expect before any solving is done?
Predict first: does y = x² + 4x + 7 reach the x-axis? How could you tell without drawing it? Then use the figure to test your idea on other quadratics.
y = x² − 4x + 3: Δ = (−4)² − 4(1)(3) = 4. Δ is positive, so there are two real solutions, x = 1 and x = 3. Δ is a perfect square, so it factorizes.
Show me the steps
- Set a = 1, b = 4 and c = 7. Δ = 16 − 28 = −12, which is negative. The parabola floats above the x-axis, so there are no real solutions.
- Keep a = 1 and b = 4 and slide c down to 4. Δ = 16 − 16 = 0. The vertex rests on the axis: one repeated solution, x = −2.
- Slide c to 3. Δ = 16 − 12 = 4, a perfect square. There are two solutions, −3 and −1, and it factorizes as (x + 1)(x + 3).
- Slide c to 1. Δ = 16 − 4 = 12, which is positive but not a perfect square. There are two solutions, −2 ± √3 (about −3.73 and −0.27), and no whole-number factorization.
The discriminant of ax² + bx + c = 0 is Δ = b² − 4ac, the part of the formula under the root.
Δ > 0: two different real solutions (the parabola crosses the x-axis twice). Δ = 0: one repeated solution (it touches the axis). Δ < 0: no real solutions (it misses the axis). When a, b and c are whole numbers and Δ is a perfect square, the solutions are rational and the quadratic factorizes.
Choosing a method: look for a common factor or a pair first; if Δ is a perfect square factorizing will work; otherwise use the formula or complete the square.
Two common mistakes. Losing a sign: for 4x² − 6x − 1, b² = (−6)² = 36, not −36, and −4ac = −4(4)(−1) = +16, so Δ = 52. And reading “no real solutions” as a mistake: it is a true answer, and in a model it can mean a target is never reached.
Worked examples
1. How many real solutions has x² − 6x + 9 = 0?
- a = 1, b = −6, c = 9: Δ = (−6)² − 4(1)(9) = 36 − 36 = 0.
- Δ = 0, so there is one repeated solution: x = 6 ÷ 2 = 3. Indeed (x − 3)² = 0.
2. Choose a method for 3x² + 8x − 3 = 0 and solve it.
- Δ = 8² − 4(3)(−3) = 64 + 36 = 100 = 10², a perfect square, so it factorizes.
- (3x − 1)(x + 3) = 0, so x = 1/3 or x = −3.
3. A rectangle has a perimeter of 24 m and an area of A m². What is the greatest possible area?
- Width w, length 12 − w, so w(12 − w) = A, which is w² − 12w + A = 0.
- A solution exists only when Δ ≥ 0: 144 − 4A ≥ 0, so A ≤ 36.
- The greatest area is 36 m², when Δ = 0 and w = 6, a square.
Your go. The first step is given. For what value of k does x² + 8x + k = 0 have exactly one solution?
- Write the discriminant in terms of k: Δ = 8² − 4(1)(k) = 64 − 4k.
- One solution means Δ = 0. Solve for k.
Check the answer
64 − 4k = 0 gives k = 16. Then x² + 8x + 16 = (x + 4)², with the single solution x = −4.
With a GDC (optional)
A GDC can show the number of solutions by graphing. Enter y = x² − 5x + 8 and look at the graph: it stays above the x-axis, so there are no real solutions, which fits Δ = 25 − 32 = −7. The zero tool reports that no zero is found. Some solvers return numbers with an i, called complex numbers; they are not real solutions and are beyond this unit.
Work out Δ by hand first, with brackets for negative values, and use the graph to check. Everything here can be done by hand.
Check before you go on
1. How many real solutions does 2x² − 3x + 9 = 0 have?
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. Find the discriminant of 3x² − 5x − 12.
3. x² + kx + 9 = 0 has exactly one solution, and k is positive. Find k.
Level 2 · Building
4. x² − 6x + k = 0 has two different real solutions when k is less than K. Find K.
5. A gardener in Hanoi has 28 m of fencing for a rectangular plot. If the width is w metres, the area A satisfies w² − 14w + A = 0. What is the greatest area, in square metres, for which a plot exists?
Level 3 · Stretch
6. y = 2x² − 16x + c touches the x-axis at exactly one point. Find c.
7. x² + 2kx + 25 = 0 has no real solutions when k lies between −m and m. Find m.
Something to think about
If a and c have opposite signs, how many real solutions does ax² + bx + c = 0 have? Try a = 1, c = −3 and a = −2, c = 5 with b = 0, then explain why it always works.
One way to think about it
Always two. If a and c have opposite signs, then ac is negative, so −4ac is positive, and Δ = b² − 4ac is a square (never negative) plus a positive number, so it is positive. With b = 0, a = 1 and c = −3, Δ = 12; with a = −2 and c = 5, Δ = 40. A parabola that opens up with its y-intercept below the axis must cross the axis twice.
Exit ticket
8. Which of these equations has rational solutions, so that it can be solved by factorizing?