Chapter 5 of 8

All 8 chapters
  1. 1Expanding two brackets, readunderstood
  2. 2Factorising quadratics, readunderstood
  3. 3Using factorising, readunderstood
  4. 4Completing the square, readunderstood
  5. 5The quadratic formula, readunderstood
  6. 6Graphs of quadratics, readunderstood
  7. 7Quadratic sequences, readunderstood
  8. 8A line and a curve, readunderstood

M4.9.5 Chapter 5

The quadratic formula

x² + 4x + 1 = 0 does not factorise with whole numbers. Can I still solve it every time?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve any quadratic equation with the formula
  • Use b² − 4ac to say how many solutions there are
  • Form and solve a quadratic from an area problem

Some quadratics do not factorise with whole numbers. Completing the square always works, but it takes many steps each time. The quadratic formula does all of that work once, for every quadratic.

Write any quadratic as ax² + bx + c = 0. The number a is in front of x², the number b is in front of x, and c stands alone. These are the coefficients. Read each one with its sign.

A table with the heads quadratic and a, b, c. Rows: x² + 4x + 1 with 1, 4, 1; 2x² − 5x + 3 with 2, −5, 3; x² − x − 6 with 1, −1, −6.quadratica, b, cx² + 4x + 11, 4, 12x² − 5x + 32, −5, 3x² − x − 61, −1, −6
A letter with no number in front, like x, has the number 1. A minus sign belongs to the number after it.

The solutions of ax² + bx + c = 0 are x = (−b ± √(b² − 4ac)) ÷ (2a). The sign ± means two answers: one with a plus and one with a minus. The whole top is divided by 2a. Here is why the formula is true.

Draw it first

Where the formula comes from, part one

  1. A column of equations with arrows between them. The first is ax² + bx + c = 0. Step 2 adds an arrow with the note multiply by 4a, and 4a²x² + 4abx + 4ac = 0. Step 3 adds an arrow with the note add b², take 4ac, and 4a²x² + 4abx + b² = b² − 4ac. Step 4 adds an arrow with the note the left is a square, and (2ax + b)² = b² − 4ac.ax² + bx + c = 0

    1 Start with any quadratic, ax² + bx + c = 0, where a is not 0.

  2. A column of equations with arrows between them. The first is ax² + bx + c = 0. Step 2 adds an arrow with the note multiply by 4a, and 4a²x² + 4abx + 4ac = 0. Step 3 adds an arrow with the note add b², take 4ac, and 4a²x² + 4abx + b² = b² − 4ac. Step 4 adds an arrow with the note the left is a square, and (2ax + b)² = b² − 4ac.ax² + bx + c = 0ax² + bx + c = 04a²x² + 4abx + 4ac = 0multiply by 4a

    2 Multiply both sides by 4a. Every term is multiplied, so the equation becomes 4a²x² + 4abx + 4ac = 0.

  3. A column of equations with arrows between them. The first is ax² + bx + c = 0. Step 2 adds an arrow with the note multiply by 4a, and 4a²x² + 4abx + 4ac = 0. Step 3 adds an arrow with the note add b², take 4ac, and 4a²x² + 4abx + b² = b² − 4ac. Step 4 adds an arrow with the note the left is a square, and (2ax + b)² = b² − 4ac.ax² + bx + c = 0ax² + bx + c = 04a²x² + 4abx + 4ac = 0multiply by 4aax² + bx + c = 04a²x² + 4abx + 4ac = 04a²x² + 4abx + b² = b² − 4acmultiply by 4aadd b², take 4ac

    3 Add b² to both sides, then take away 4ac from both sides. The left side is now 4a²x² + 4abx + b², and the right side is b² − 4ac.

  4. A column of equations with arrows between them. The first is ax² + bx + c = 0. Step 2 adds an arrow with the note multiply by 4a, and 4a²x² + 4abx + 4ac = 0. Step 3 adds an arrow with the note add b², take 4ac, and 4a²x² + 4abx + b² = b² − 4ac. Step 4 adds an arrow with the note the left is a square, and (2ax + b)² = b² − 4ac.ax² + bx + c = 0ax² + bx + c = 04a²x² + 4abx + 4ac = 0multiply by 4aax² + bx + c = 04a²x² + 4abx + 4ac = 04a²x² + 4abx + b² = b² − 4acmultiply by 4aadd b², take 4acax² + bx + c = 04a²x² + 4abx + 4ac = 04a²x² + 4abx + b² = b² − 4ac(2ax + b)² = b² − 4acmultiply by 4aadd b², take 4acthe left is a square

    4 The left side is a perfect square. Check it with a grid as in chapter 1: (2ax + b)(2ax + b) = 4a²x² + 2abx + 2abx + b². So (2ax + b)² = b² − 4ac.

Now take the square root of both sides. This works only when b² − 4ac is 0 or more, because no real number squared is below 0. A positive number has two square roots, so write ±.

Draw it first

Where the formula comes from, part two

  1. A column of equations with arrows between them. The first is (2ax + b)² = b² − 4ac. Step 2 adds an arrow with the note take the square root, and 2ax + b = ± √(b² − 4ac). Step 3 adds an arrow with the note take b, divide by 2a, and x = (−b ± √(b² − 4ac)) ÷ (2a).(2ax + b)² = b² − 4ac

    1 We have (2ax + b)² = b² − 4ac.

  2. A column of equations with arrows between them. The first is (2ax + b)² = b² − 4ac. Step 2 adds an arrow with the note take the square root, and 2ax + b = ± √(b² − 4ac). Step 3 adds an arrow with the note take b, divide by 2a, and x = (−b ± √(b² − 4ac)) ÷ (2a).(2ax + b)² = b² − 4ac(2ax + b)² = b² − 4ac2ax + b = ± √(b² − 4ac)take the square root

    2 Take the square root of both sides, which needs b² − 4ac to be 0 or more. There are two square roots, so write ±.

  3. A column of equations with arrows between them. The first is (2ax + b)² = b² − 4ac. Step 2 adds an arrow with the note take the square root, and 2ax + b = ± √(b² − 4ac). Step 3 adds an arrow with the note take b, divide by 2a, and x = (−b ± √(b² − 4ac)) ÷ (2a).(2ax + b)² = b² − 4ac(2ax + b)² = b² − 4ac2ax + b = ± √(b² − 4ac)take the square root(2ax + b)² = b² − 4ac2ax + b = ± √(b² − 4ac)x = (−b ± √(b² − 4ac)) ÷ (2a)take the square roottake b, divide by 2a

    3 Take b from both sides and divide by 2a. That is the formula.

To use the formula, first read off a, b and c. Then work out the part under the root, b² − 4ac, on its own. Last, put everything in.

Worked example

Solve x² + 3x − 10 = 0 with the formula.

  1. Read off the numbers: a = 1, b = 3 and c = −10.
  2. Work out b² − 4ac = 3 × 3 − 4 × 1 × (−10) = 9 + 40 = 49.
  3. The square root of 49 is 7.
  4. Use the formula: x = (−3 + 7)/2 or x = (−3 − 7)/2.
  5. So x = 4/2 = 2 or x = −10/2 = −5.
  6. Check with x = 2: 2 × 2 + 3 × 2 − 10 = 4 + 6 − 10 = 0.

Answer: x = 2 or x = −5

This quadratic also factorises: (x + 5)(x − 2) = 0 gives the same answers. The formula also works when factorising does not, and it is quicker than completing the square.

Take care when b is negative. The formula starts with −b, and −(−4) is +4. The part b² is the square of b, so (−4)² = 16, which is positive.

Worked example

Solve x² − 4x − 3 = 0. Give the answers to 2 decimal places.

  1. Read off the numbers: a = 1, b = −4 and c = −3.
  2. Work out b² − 4ac = (−4) × (−4) − 4 × 1 × (−3) = 16 + 12 = 28.
  3. The square root of 28 is about 5.2915.
  4. Use the formula. Since b = −4, −b = 4. So x = (4 + 5.2915)/2 or x = (4 − 5.2915)/2.
  5. Work out each one: x = 9.2915/2 ≈ 4.65, or x = −1.2915/2 ≈ −0.65.

Answer: x ≈ 4.65 or x ≈ −0.65

In every example so far a = 1, so the bottom of the formula was 2. When a is not 1, the bottom is 2a, which changes. Here is one with a = 2.

Worked example

Solve 2x² + 3x − 2 = 0 with the formula.

  1. Read off the numbers: a = 2, b = 3 and c = −2.
  2. Work out b² − 4ac = 9 − 4 × 2 × (−2) = 9 + 16 = 25.
  3. The square root of 25 is 5.
  4. The bottom of the formula is 2a, and 2a = 2 × 2 = 4.
  5. So x = (−3 + 5)/4 or x = (−3 − 5)/4.
  6. Work out each one: x = 2/4 = 0.5 or x = −8/4 = −2.
  7. Check with x = 0.5: 2 × 0.25 + 1.5 − 2 = 0.

Answer: x = 0.5 or x = −2

The part under the root is called the discriminant, D = b² − 4ac. It tells you how many solutions there are before you finish. Take three quadratics that differ only in the last number.

A table with the heads equation, D and answers. Rows: x² − 4x + 3 with D 4 and 2 solutions; x² − 4x + 4 with D 0 and 1 solution; x² − 4x + 5 with D −4 and 0 solutions.equationDanswersx² − 4x + 342x² − 4x + 401x² − 4x + 5−40
D above 0 gives two solutions. D equal to 0 gives one. D below 0 gives none.

Here are their graphs. A solution is a place where the curve meets the x axis. First D = 4. The square root of 4 is 2, so the formula gives two different answers.

A graph with the x axis x and the y axis y. A curve named y = x² − 4x + 3 crosses the x axis at two points, A and B.ABy = x² − 4x + 31345−2−112345670xy
D = 4. The curve crosses the x axis twice, at x = 1 and x = 3.

When D = 0, the square root is 0, so plus and minus give the same answer. The curve just touches the x axis.

A graph with the x axis x and the y axis y. A curve named y = x² − 4x + 4 touches the x axis at one point, A.Ay = x² − 4x + 412345−2−112345670xy
D = 0. The curve touches the x axis once, at x = 2.

When D is below 0, the formula needs the square root of a negative number. No real number squared makes a negative, so there is no real solution. The curve stays above the x axis.

A graph with the x axis x and the y axis y. A curve named y = x² − 4x + 5 stays above the x axis and never meets it.y = x² − 4x + 512345−2−112345670xy
D = −4. The curve never meets the x axis, so the equation has no solution.

Worked example

How many solutions does 2x² + 3x + 5 = 0 have?

  1. Read off the numbers: a = 2, b = 3 and c = 5.
  2. Work out b² − 4ac = 3 × 3 − 4 × 2 × 5 = 9 − 40 = −31.
  3. The discriminant is below 0, so the formula needs the square root of a negative number. There is no such real number, so there is no real solution.

Answer: No real solution. The discriminant is −31.

Quadratic equations come from area problems. A rectangle is 3 m longer than it is wide, and its area is 28 m². Call the width x metres. Then the length is x + 3 metres.

A rectangle with its width marked x m on the left and its length marked (x + 3) m along the bottom. The area, 28 m², is in the text.(x + 3) mx m
Area is length times width, so x(x + 3) = 28.

Worked example

A rectangle is 3 m longer than it is wide. Its area is 28 m². How long and wide is it?

  1. Let x be the width in metres. The length is x + 3 metres.
  2. Area = length × width, so x(x + 3) = 28.
  3. Expand and make one side 0: x² + 3x − 28 = 0.
  4. Use the formula with a = 1, b = 3 and c = −28. Then b² − 4ac = 9 + 112 = 121, and the square root of 121 is 11.
  5. So x = (−3 + 11)/2 = 4, or x = (−3 − 11)/2 = −7.
  6. A width cannot be negative, so x = 4. The rectangle is 4 m wide and 7 m long.
  7. Check the area: 4 × 7 = 28.

Answer: The rectangle is 4 m wide and 7 m long.

See it three ways

With things

Draw the curve y = x² − 6x + 5 from a table of values. Read where it crosses the x axis.

As a picture

A graph with the x axis x and the y axis y. A curve named y = x² − 6x + 5 crosses the x axis at two points, A and B.ABy = x² − 6x + 5123456−5−4−3−2−11234560xy

In words

The curve crosses the x axis at 1 and 5, so those are the solutions. The formula gives the same: b² − 4ac = 36 − 20 = 16, and the square root of 16 is 4.

In symbols

x = (6 ± 4)/2, so x = 5 or x = 1.

Try it

Solve x² + 4x + 1 = 0 and test it

You need: paper, a pencil and a calculator.

  1. Read off a = 1, b = 4 and c = 1. Work out b² − 4ac = 16 − 4 = 12.
  2. Use the formula with √12 ≈ 3.4641: x = (−4 + 3.4641)/2 ≈ −0.27, or x = (−4 − 3.4641)/2 ≈ −3.73.
  3. Put each answer back into x² + 4x + 1 and write what you get.

What you will see: Both answers give a value very close to 0. They are not exactly 0 only because the answers are rounded.

If you cannot try it, this is what happens.
A table with the heads x and x² + 4x + 1. Rows: x is −0.27 and the value is −0.0071; x is −3.73 and the value is −0.0071.xx² + 4x + 1−0.27−0.0071−3.73−0.0071

A wrong idea many people have

When b = −6, b² = −36.

Why it feels right: The minus sign seems to stay in front of the number, so the square looks negative too.

The test that breaks it: Work out (−6) × (−6). A negative times a negative is positive, so it is 36.

What is true: The square of a negative number is positive. In the formula, b² is the square of the whole of b, sign included. With b = −6, b² = 36.

A wrong idea many people have

x = −b ± √(b² − 4ac) / 2a: only the root is divided by 2a.

Why it feels right: Written on one line, the division seems to belong to the last part only.

The test that breaks it: Take x² − 2x − 8 = 0. Dividing only the root gives 2 ± 6/2 = 2 ± 3, so x = 5. But 5 × 5 − 2 × 5 − 8 = 7, not 0.

What is true: The whole top, −b ± √(b² − 4ac), is divided by 2a. For this equation x = (2 ± 6)/2, so x = 4 or x = −2.

A wrong idea many people have

A quadratic always has two solutions, because of the ±.

Why it feels right: The formula has a plus or minus, so it seems that every quadratic gives two answers.

The test that breaks it: Try x² + 1 = 0. This needs x² = −1, but no real number squared makes −1. The formula has D = −4, with no real square root.

What is true: The ± gives two answers only when the discriminant is above 0. When D = 0 there is one answer, and when D is below 0 there is no real solution.

Key words

coefficient
the number in front of a letter, such as the 5 in 5x, with its sign.
discriminant
the number b² − 4ac, which tells you how many solutions a quadratic has.

Back to your guess

The answer is A: Yes. One formula uses the three numbers in the equation.

One formula solves every quadratic. It comes from completing the square on the letters a, b and c, so it works for any numbers. Trying whole numbers is slow, and it misses answers that are not whole.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. What are a, b and c in 3x² − 4x + 7 = 0?

    Tap your answer.

    1. Not this one. The x term is −4x, so b = −4. A minus sign belongs to the number after it. Go to M4.9.5. Try another.

    2. Yes, that is right. The number in front of x² is 3. The number in front of x is −4. The number on its own is 7.

    3. Not this one. The last number is +7, with no minus sign in front of it. So c = 7. Go to M4.9.5. Try another.

    4. Not this one. The number a is the number in front of x², which is 3. Go to M4.9.5. Try another.

    See the answer

    a = 3, b = −4, c = 7

    The number in front of x² is 3. The number in front of x is −4. The number on its own is 7.

    • If you chose A: The x term is −4x, so b = −4. A minus sign belongs to the number after it. Go to M4.9.5.
    • If you chose C: The last number is +7, with no minus sign in front of it. So c = 7. Go to M4.9.5.
    • If you chose D: The number a is the number in front of x², which is 3. Go to M4.9.5.

  2. How many solutions does x² + 2x + 5 = 0 have?

    Tap your answer.

    1. Not this one. The ± gives two answers only when the part under the root is positive. Here b² − 4ac = 4 − 20 = −16, which is below 0. Go to M4.9.5. Try another.

    2. Not this one. There is one solution only when b² − 4ac is exactly 0. Here it is 4 − 20 = −16. Go to M4.9.5. Try another.

    3. Yes, that is right. Here a = 1, b = 2 and c = 5. So b² − 4ac = 4 − 20 = −16. No real number squared makes −16, so there is no real solution.

    See the answer

    None. The discriminant is −16, below zero.

    Here a = 1, b = 2 and c = 5. So b² − 4ac = 4 − 20 = −16. No real number squared makes −16, so there is no real solution.

    • If you chose A: The ± gives two answers only when the part under the root is positive. Here b² − 4ac = 4 − 20 = −16, which is below 0. Go to M4.9.5.
    • If you chose B: There is one solution only when b² − 4ac is exactly 0. Here it is 4 − 20 = −16. Go to M4.9.5.

  3. How many solutions does x² − 6x + 9 = 0 have?

    Tap your answer.

    1. Yes, that is right. Here a = 1, b = −6 and c = 9. So b² − 4ac = 36 − 36 = 0. The formula gives x = 6/2 = 3, a single answer.

    2. Not this one. When b² − 4ac = 0, the root is 0, so + and − give the same answer. Here b² − 4ac = 36 − 36 = 0. Go to M4.9.5. Try another.

    3. Not this one. There is no solution only when b² − 4ac is below 0. Here it is 0, and 0 has a square root, 0. Go to M4.9.5. Try another.

    See the answer

    One. The discriminant is 0.

    Here a = 1, b = −6 and c = 9. So b² − 4ac = 36 − 36 = 0. The formula gives x = 6/2 = 3, a single answer.

    • If you chose B: When b² − 4ac = 0, the root is 0, so + and − give the same answer. Here b² − 4ac = 36 − 36 = 0. Go to M4.9.5.
    • If you chose C: There is no solution only when b² − 4ac is below 0. Here it is 0, and 0 has a square root, 0. Go to M4.9.5.

  4. Which equation has two different solutions?

    Tap your answer.

    1. Not this one. Here b² − 4ac = 16 − 16 = 0, so there is only one solution. Go to M4.9.5. Try another.

    2. Not this one. Here b² − 4ac = 16 − 36 = −20, which is below 0, so there is no solution. Go to M4.9.5. Try another.

    3. Not this one. Here b² − 4ac = 36 − 40 = −4, which is below 0, so there is no solution. Go to M4.9.5. Try another.

    4. Yes, that is right. Here b² − 4ac = 16 − 8 = 8, which is above 0, so there are two solutions.

    See the answer

    x² + 4x + 2 = 0

    Here b² − 4ac = 16 − 8 = 8, which is above 0, so there are two solutions.

    • If you chose A: Here b² − 4ac = 16 − 16 = 0, so there is only one solution. Go to M4.9.5.
    • If you chose B: Here b² − 4ac = 16 − 36 = −20, which is below 0, so there is no solution. Go to M4.9.5.
    • If you chose C: Here b² − 4ac = 36 − 40 = −4, which is below 0, so there is no solution. Go to M4.9.5.

  5. Solve x² − 2x − 15 = 0 with the formula.

    Tap your answer.

    1. Not this one. The formula starts with −b. Here b = −2, so −b = +2, not −2. Go to M4.9.5. Try another.

    2. Yes, that is right. Here b² − 4ac = 4 + 60 = 64 and its square root is 8. So x = (2 + 8)/2 = 5 or x = (2 − 8)/2 = −3.

    3. Not this one. You divided only the root by 2, which gives 2 ± 4. The whole top, 2 ± 8, is divided by 2a. Go to M4.9.5. Try another.

    See the answer

    x = 5 or x = −3

    Here b² − 4ac = 4 + 60 = 64 and its square root is 8. So x = (2 + 8)/2 = 5 or x = (2 − 8)/2 = −3.

    • If you chose A: The formula starts with −b. Here b = −2, so −b = +2, not −2. Go to M4.9.5.
    • If you chose C: You divided only the root by 2, which gives 2 ± 4. The whole top, 2 ± 8, is divided by 2a. Go to M4.9.5.

  6. Solve 2x² − 5x + 3 = 0 with the formula.

    Tap your answer.

    1. Yes, that is right. Here b² − 4ac = 25 − 24 = 1. So x = (5 ± 1)/4, which is 1.5 or 1.

    2. Not this one. You divided by 2. The bottom of the formula is 2a, and 2a = 4 here. Go to M4.9.5. Try another.

    3. Not this one. The formula starts with −b. Here b = −5, so −b = +5, not −5. Go to M4.9.5. Try another.

    See the answer

    x = 1.5 or x = 1

    Here b² − 4ac = 25 − 24 = 1. So x = (5 ± 1)/4, which is 1.5 or 1.

    • If you chose B: You divided by 2. The bottom of the formula is 2a, and 2a = 4 here. Go to M4.9.5.
    • If you chose C: The formula starts with −b. Here b = −5, so −b = +5, not −5. Go to M4.9.5.

  7. A rectangle is 5 m longer than it is wide, and its area is 24 m². Let x be the width in metres. Which equation is right?

    Tap your answer.

    1. Not this one. The length is x + 5, not 5x. And area is a product, width × length, not a sum. Go to M4.9.5. Try another.

    2. Not this one. The area is x(x + 5). Expanding gives x² + 5x, not x² + 5. Go to M4.9.1. Try another.

    3. Yes, that is right. The length is x + 5 and the width is x. Area = x(x + 5) = x² + 5x, and the area is 24.

    4. Not this one. That adds the four sides, which is the perimeter. Area is length times width. Go to M4.9.5. Try another.

    See the answer

    x² + 5x = 24

    The length is x + 5 and the width is x. Area = x(x + 5) = x² + 5x, and the area is 24.

    • If you chose A: The length is x + 5, not 5x. And area is a product, width × length, not a sum. Go to M4.9.5.
    • If you chose B: The area is x(x + 5). Expanding gives x² + 5x, not x² + 5. Go to M4.9.1.
    • If you chose D: That adds the four sides, which is the perimeter. Area is length times width. Go to M4.9.5.

  8. A rectangle is 5 m longer than it is wide, with area 24 m². Its width x m solves x² + 5x − 24 = 0, which has the answers x = 3 and x = −8. What is the width?

    Tap your answer.

    1. Not this one. A width is a length, and a length cannot be negative. Go to M4.9.5. Try another.

    2. Not this one. Only the positive answer can be a width. The negative answer does not fit the problem. Go to M4.9.5. Try another.

    3. Not this one. The number 8 is the length, since x + 5 = 8. The width is x. Go to M4.9.5. Try another.

    4. Yes, that is right. The width cannot be negative, so x = 3. The length is 3 + 5 = 8 m, and 3 × 8 = 24.

    See the answer

    x = 3 m

    The width cannot be negative, so x = 3. The length is 3 + 5 = 8 m, and 3 × 8 = 24.

    • If you chose A: A width is a length, and a length cannot be negative. Go to M4.9.5.
    • If you chose B: Only the positive answer can be a width. The negative answer does not fit the problem. Go to M4.9.5.
    • If you chose C: The number 8 is the length, since x + 5 = 8. The width is x. Go to M4.9.5.

  9. From an earlier chapter

    Solve (x + 1)² = 16.

    Tap your answer.

    1. Yes, that is right. Take the square root of both sides: x + 1 = 4 or x + 1 = −4. Take 1 from each side: x = 3 or x = −5.

    2. Not this one. 16 has two square roots, 4 and −4. Then x + 1 = −4 gives x = −5 as well. Go to M4.9.4. Try another.

    3. Not this one. You took x + 1 = 16, which skips the square root. First take the square root of both sides. Go to M4.9.4. Try another.

    4. Not this one. Take 1 from both sides: x + 1 = 4 gives x = 3, not −3. The 1 is taken away. Go to M4.9.4. Try another.

    See the answer

    x = 3 or x = −5

    Take the square root of both sides: x + 1 = 4 or x + 1 = −4. Take 1 from each side: x = 3 or x = −5.

    • If you chose B: 16 has two square roots, 4 and −4. Then x + 1 = −4 gives x = −5 as well. Go to M4.9.4.
    • If you chose C: You took x + 1 = 16, which skips the square root. First take the square root of both sides. Go to M4.9.4.
    • If you chose D: Take 1 from both sides: x + 1 = 4 gives x = 3, not −3. The 1 is taken away. Go to M4.9.4.

  10. The length of a rectangle is 3 m more than its width. Its area is 40 m². How wide is it, in metres?

    See the answer

    5 m

    Let x be the width. Then x(x + 3) = 40, so x² + 3x − 40 = 0. The formula gives x = (−3 ± 13)/2, so x = 5 or x = −8. The width is 5 m.

  11. Work out b² − 4ac for 2x² + 3x − 5 = 0.

    See the answer

    49

    Here a = 2, b = 3 and c = −5. So b² − 4ac = 9 − 4 × 2 × (−5) = 9 + 40 = 49.

Keep it

Copy this into your exercise book.

Read off a, b and c with their signs. Work out b² − 4ac first. Then x = (−b ± √(b² − 4ac)) ÷ (2a). Above 0: two answers. At 0: one. Below 0: no real answer.

Draw: Three curves: one crossing the x axis twice, one touching it, one above it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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