Year 11 Extended · Unit 6 Logarithms and non-linear inequalities · Lesson 7 of 8

For which values of x is a quadratic positive or negative?

A quadratic changes sign only at its roots. This lesson finds the roots, draws a sign diagram, and reads off the set of x-values where a quadratic is above, below or on zero.

Number and AlgebraNo calculator

Warm-up

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

  1. From Year 9: factorize x2 + 4x − 21.
    Answer(x + 7)(x − 3).
  2. From Year 10: solve x2 − 9x + 20 = 0.
    Answerx = 4 or x = 5.
  3. From Year 8: solve 5 − 2x > 11.
    Answerx < −3. Dividing by −2 reverses the inequality.
Words used in this lesson
inequality
a statement that one expression is less than or greater than another. For example: x2 < 10.
critical value
a number where an expression changes sign. For example: a root of the quadratic.
sign diagram
a number line showing where an expression is positive or negative. For example: plus, minus, plus.
strict inequality
one that uses < or > and excludes the end values. For example: x > 5.
discriminant
b2 − 4ac, which tells the number of real roots. For example: negative: no real roots.

For which numbers of items is a stall in profit?

A market stall in Jaipur sells x hundred cups of tea a day. After the cost of the tea, the stall and the helper, the daily profit in thousands of rupees is P = −x2 + 8x − 12. At x = 0 the profit is negative, because the costs have to be paid whatever is sold.

Predict first: between which numbers of hundreds of cups will the stall be in profit, and what number of cups do you think makes the most profit? Then use the figure to see how the roots and the direction of the parabola decide the signs.

−2
5
up
x < −2 or x > 3y is positive when
−2 < x < 3y is negative when

y = (x + 2)(x − 3): positive when x < −2 or x > 3, negative when −2 < x < 3.

Show me the steps
  1. y = (x + 4)(x − 2), opening up: positive when x < −4 or x > 2, negative when −4 < x < 2.
  2. y = −x(x − 5), opening down: positive when 0 < x < 5, negative when x < 0 or x > 5.
  3. y = (x − 2)(x − 3), opening up: positive when x < 2 or x > 3, negative when 2 < x < 3.
  4. Roots p and q with p below q: opening up, y is positive outside the roots and negative between them. Opening down, the two regions swap.

A quadratic can only change sign at a root. To solve an inequality such as ax2 + bx + c > 0:

  1. Move everything to one side so that the other side is 0.
  2. Find the roots by factorizing or by the formula. They are the critical values.
  3. Draw a sketch or a sign diagram. If a is positive the parabola is above zero outside the roots and below zero between them; if a is negative the regions swap.
  4. Write the solution, using < or > for strict inequalities and ≤ or ≥ when the roots are included.

Two common slips. Taking a square root and keeping only one side: x2 > 49 is true for x = −8 as well as x = 8, so the solution is x < −7 or x > 7. And dividing both sides by x: from x2 < 4x, dividing by x gives x < 4, which wrongly allows every negative x (try x = −2: 4 < −8 is false).

Worked examples

1. Solve x2 − x − 12 < 0.

  1. Factorize: (x − 4)(x + 3) < 0, with roots 4 and −3.
  2. The parabola opens up, so it is below zero between the roots: −3 < x < 4.

2. Solve 2x2 + 5x ≥ 3.

  1. Move the 3: 2x2 + 5x − 3 ≥ 0, which factorizes as (2x − 1)(x + 3) ≥ 0, with roots 0.5 and −3.
  2. The parabola is above zero outside the roots and zero at them: x ≤ −3 or x ≥ 0.5.

3. Solve 10 + 3x − x2 ≥ 0.

  1. The x2 term is negative, so multiply by −1 and reverse the inequality: x2 − 3x − 10 ≤ 0, which is (x − 5)(x + 2) ≤ 0.
  2. Below zero between the roots, and zero at them: −2 ≤ x ≤ 5.

Your go. The first step is given. Solve x2 + 2x − 15 > 0. It factorizes as (x + 5)(x − 3).

  1. The roots are −5 and 3, and the parabola opens upwards.
  2. Decide whether > 0 means outside or between the roots.
Check the answer

Above zero is outside the roots: x < −5 or x > 3.

Check before you go on

1. Which describes the solution of x2 − 10x + 21 > 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. How many integers x satisfy x2 − 4x − 5 < 0?

3. Find the largest integer x with x2 + 3x − 28 ≤ 0.

Level 2 · Building

4. How many integers x from −3 to 10 inclusive satisfy x2 > 7x?

5. A rectangular garden is x metres wide and (x + 5) metres long. Its area must be less than 36 m2. What is the largest whole number of metres the width can be?

Level 3 · Stretch

6. How many integer values of k make x2 + kx + 9 = 0 have no real roots?

7. How many integers x satisfy both x2 − 6x + 5 ≥ 0 and x2 < 16?

Something to think about

A quadratic with no real roots, such as x2 + 4, is always above the axis. What does that say about the solution of x2 + 4 < 0, and of x2 + 4 > 0?

One way to think about it

x2 + 4 is at least 4 for every x, so x2 + 4 < 0 has no solution at all, and x2 + 4 > 0 is true for every x. With no roots the sign never changes, so the whole axis has one sign.

Exit ticket

8. Which describes the solution of 24 + 2x − x2 > 0?