Year 10 Standard · Unit 3 Functions and quadratics · Lesson 4 of 6

What do you know about two numbers whose product is zero?

Some quadratic equations can be split into two simpler equations and solved one piece at a time. This lesson finds out when that is allowed, and why.

FunctionsGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 10, Lesson 2: factorize x² + 2x − 15.
    Answer(x + 5)(x − 3). The numbers have product −15 and sum 2.
  2. From Year 8: solve 3x − 12 = 0.
    Answerx = 4. Add 12 to both sides, then divide by 3.
  3. Work out (−5) × 0.
    Answer0. Anything multiplied by 0 is 0.
Words used in this lesson
null factor law
if a product is zero, one of its factors is zero. For example: if mn = 0 then m = 0 or n = 0.
solution
a value of the unknown that makes the equation true. For example: x = 2 for x + 1 = 3.
root
another name for a solution, or a zero of a function. For example: the roots of x² − 1 = 0.
rearrange
move terms so one side is zero. For example: x² = 5x becomes x² − 5x = 0.
repeated root
the same solution twice, from a squared factor. For example: x = 1 in (x − 1)² = 0.
reject
decide a solution does not fit the situation. For example: a negative length.

A product of zero

Take two numbers and multiply them. If the answer is 0, what do you know about the two numbers? Try a few: 0 × 7, 5 × 0, 0 × 0. Can you find two numbers that are both not zero and still multiply to give 0?

Predict first: for (x − 6)(x + 9) = 0, which values of x are possible? Use the figure to see where a product of brackets is zero.

−2
3
0
2first bracket
−3second bracket
−6product

y = (x + 2)(x − 3) at x = 0: the brackets are 2 and −3, and the product is −6. No bracket is 0, so the product is not 0. The product is 0 only at x = −2 and x = 3.

Show me the steps
  1. With p = −2 and q = 3, slide x to 3. The second bracket is 3 − 3 = 0, so the product is 0 whatever the first bracket is (it is 5). The point sits on the x-axis.
  2. Slide x to 1. The brackets are 3 and −2, and the product is −6. Neither bracket is 0, so the point is off the axis.
  3. Set p = 4 and q = 4. Both brackets are zero at the same x = 4, so there is only one solution and the parabola just touches the axis.
  4. Set p = 0 and q = 5. The first bracket is just x, so y = x(x − 5): the solutions are x = 0 and x = 5. The solution x = 0 is the one students most often lose.

The null factor law: if a × b = 0 then a = 0 or b = 0 (or both). So to solve a quadratic equation:

  1. Rearrange it so one side is 0.
  2. Factorize the other side.
  3. Set each factor equal to 0.
  4. Solve each.
  5. Check both in the original equation.

The solutions are where the graph of the quadratic meets the x-axis.

Two common mistakes. The law only works for zero: (x − 3)(x + 2) = 10 does not mean x − 3 = 10 or x + 2 = 10, because many pairs of numbers multiply to 10. And dividing both sides of x² = 8x by x loses the solution x = 0; move everything to one side and factorize instead.

Worked examples

1. Solve x² − 3x − 10 = 0.

  1. One side is already 0. Factorize: the numbers with product −10 and sum −3 are −5 and 2.
  2. (x − 5)(x + 2) = 0, so x − 5 = 0 or x + 2 = 0.
  3. x = 5 or x = −2. Check x = −2: 4 + 6 − 10 = 0.

2. Solve x² = 8x.

  1. Make one side 0: x² − 8x = 0. Factorize by taking out x: x(x − 8) = 0.
  2. x = 0 or x − 8 = 0, so x = 0 or x = 8. Both work: 0 = 0 and 64 = 64.

3. Solve 2x² + x = 6.

  1. Subtract 6 from both sides: 2x² + x − 6 = 0.
  2. Factorize: (2x − 3)(x + 2) = 0.
  3. 2x − 3 = 0 gives x = 1.5, and x + 2 = 0 gives x = −2. So x = 1.5 or x = −2.

Your go. The first step is given. Solve x² + 4x = 21.

  1. Make one side zero: x² + 4x − 21 = 0.
  2. Factorize, set each bracket to 0, and solve. Then check both.
Check the answer

(x + 7)(x − 3) = 0, so x = −7 or x = 3. Check x = 3: 9 + 12 = 21.

With a GDC (optional)

A GDC can check your solutions graphically. Graph y = x² − 11x + 24 and use the zero or root tool twice, once on each side of the turning point: it reports the two x-intercepts, 3 and 8, which are the solutions of x² − 11x + 24 = 0. A table of values with y = 0 shows the same thing.

Take care: an equation solver may report only one solution, so look at the graph to see whether there is another. Write down both solutions, and check each in the original equation. Everything here can be done by hand.

Check before you go on

1. Solve (x − 4)(x + 7) = 0.

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. x² − 9x + 14 = 0 has two solutions. What is the larger one?

3. Solve x² = 12x. What is the smaller solution?

Level 2 · Building

4. 2x² − 7x + 3 = 0 has two solutions. What is the smaller one?

5. A rectangular vegetable plot in Kerala is 3 m longer than it is wide, and its area is 54 m². Let the width be x metres, so x(x + 3) = 54. What is the width, in metres?

Level 3 · Stretch

6. x(x + 5) = 2x + 28 has two solutions, one positive and one negative. What is the positive solution?

7. The product of two consecutive positive whole numbers is 156. What is the larger of the two numbers?

Something to think about

Solve (x − 3)(x − 3) = 0. How many different solutions does it have, and what does the graph of y = (x − 3)² do at that solution? Compare with y = (x − 3)(x + 1).

One way to think about it

There is just one solution, x = 3, because both brackets become zero at the same time. The graph of y = (x − 3)² touches the x-axis at x = 3 and turns back without crossing, while y = (x − 3)(x + 1) crosses at two points, 3 and −1. A repeated factor means a repeated solution, a case the discriminant will pick out in Lesson 6.

Exit ticket

8. Which gives all the solutions of x² − 2x = 15?