1. HKDSE Mathematics Ronald Hui Tak Sun Secondary School

2. Summary on “AND” 22 October 2015 Ronald HUI

3. Summary on “OR” 22 October 2015 Ronald HUI

4. Book 5A Chapter 3 Solving Quadratic Inequalities in One Unknown by the Algebraic Method

5. Apart from the graphical method, are there any other methods to solve quadratic inequality? Yes, there are two common algebraic methods to solve a quadratic inequality. They are

6. Apart from the graphical method, are there any other methods to solve quadratic inequality? (1) Method of Factorization (2) Method of Tabulation

7. Note: These rules still hold when ‘>’ is replaced by ‘  ’, and ‘<‘ is replaced by ‘  ’. Method of Factorization Consider two real numbers a and b. We have the following rules: Rule 1 If ab > 0, then or. Rule 2 If ab < 0, then or. If the product of two numbers is positive, then the two numbers have the same sign. If the product of two numbers is negative, then the two numbers have different signs.

8. Let us illustrate how to solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of factorization.

9. Step 1 Factorize the quadratic expression. Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of factorization. x 2 – 7x + 10 > 0 (x – 2)(x – 5) > 0

10. Step 2 Convert the quadratic inequality into a compound inequality by the two rules. Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of factorization. In step 1, we have (x – 2)(x – 5) > 0. Then, by rule 1, we have or

11. Step 3 Solve the compound inequality. Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of factorization. In step 2, we have or. (x – 2 > 0 and x – 5 > 0) (x > 2and x > 5) (x – 2 < 0 and x – 5 < 0) (x < 2and x < 5) x < 2 ∴ The solutions of x 2 – 7x + 10 > 0 are x 5. x > 5 or

12. Follow-up question Solve the quadratic inequality x 2 + 2x – 8  0 by the method of factorization. x 2 + 2x – 8  0 (x  –4 and x  2) or (x  –4 and x  2) –4  x  2 or no solutions ∴ The solutions of x 2 + 2x – 8  0 are –4  x  2. or      02 04 x x      02 04 x x  Factorize x 2 + 2x – 8. (x + 4)(x – 2)  0  By rule 2.

13. Apart from the method of factorization, we can also solve a quadratic inequality algebraically by the method of tabulation.

14. x > 52 < x < 5 Method of Tabulation Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of tabulation. Step 1 Factorize the quadratic expression. x 2 – 7x + 10 > 0 (x – 2)(x – 5) > 0 Step 2 Divide the number line into three intervals using the roots of the corresponding quadratic equation. 2 5 x < 2 x The roots of (x – 2)(x – 5) = 0 are 2 and 5. 2 5

15. Step 3 Construct a table to determine the sign of the quadratic expression in each interval. Then, read the solutions from the table. x < 2x = 22 < x < 5x = 5x > 5 x – 2 x – 5 – 0 + ++ –– – 0 + Try x = 0. x – 2 = 0 – 2 = –2 < 0 x – 5 = 0 – 5 = –5 < 0 Try x = 3. x – 2 = 3 – 2 = 1 > 0 x – 5 = 3 – 5 = –2 < 0 Try x = 6. x – 2 = 6 – 2 = 4 > 0 x – 5 = 6 – 5 = 1 > 0 Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of tabulation. Method of Tabulation

16. x < 2x = 22 < x < 5x = 5x > 5 x – 2 x – 5 (x – 2)(x – 5) 0 + ++ – – 0 + + 0 – 0 + – – From the table, (x – 2)(x – 5) > 0 ∴ The solutions of x 2 – 7x + 10 > 0 are x 5. Step 3 Construct a table to determine the sign of the quadratic expression in each interval. Then, read the solutions from the table. Solve the quadratic inequality x 2 – 7x + 10 > 0 by the method of tabulation. Method of Tabulation x < 2 x > 5 when and.

17. x < –3x = –3–3 < x < 2x = 2x > 2 x + 3 x – 2 (x + 3)(x – 2) Follow-up question Solve the quadratic inequality x 2 + x – 6  0 by the method of tabulation. x 2 + x – 6  0 (x + 3)(x – 2)  0  Factorize x 2 + x – 6. – 0 + + + – – – 0 + + 0 – 0 + From the table, the solutions of x 2 + x – 6  0 are –3  x  2. 

18. Let me show you in the following examples. How can we solve a quadratic inequality algebraically if its corresponding quadratic equation has a double real root or no real roots?

19. Special Case (double real root) Solve the quadratic inequality x 2 – 4x + 4  0 by the algebraic method. Method 1 x 2 – 4x + 4 = (x – 2) 2 ∵ (x – 2) 2  0 holds only when x – 2 = 0, i.e. x = 2. ∴ The solution of x 2 – 4x + 4  0 is x = 2. The square of a number is always non-negative.

20. Special Case (double real root) Solve the quadratic inequality x 2 – 4x + 4  0 by the algebraic method. In fact, if the corresponding quadratic equation has a double real root, we can still use the previous two algebraic methods.

21. ∴ x = 2 Special Case (double real root) Solve the quadratic inequality x 2 – 4x + 4  0 by the algebraic method. Method 2 (method of factorization) x 2 – 4x + 4  0 x  2 and x  2 ∴ The solution of x 2 – 4x + 4  0 is x = 2.      02 02 x x  Factorize x 2 – 4x + 4. (x – 2)(x – 2)  0  By rule 2. x – 2  0 and x – 2  0

22. x < 2x = 2x > 2 x – 2 (x – 2) 2  The quadratic equation has a double real root 2. x 2 – 4x + 4 = (x – 2) 2 + 0– 0 + + Special Case (double real root) Solve the quadratic inequality x 2 – 4x + 4  0 by the algebraic method. Method 3 (method of tabulation) From the table, the solution of x 2 – 4x + 4  0 is x = 2.

23. Special Case (no real root) Solve the quadratic inequality x 2 + 2x + 3  0 by the algebraic method. However, if the corresponding quadratic equation has no real root, we cannot use the previous two algebraic methods.

24. Special Case (no real root) Solve the quadratic inequality x 2 + 2x + 3  0 by the algebraic method. For the corresponding equation x 2 + 2x + 3 = 0, = 2 2 – 4(1)(3) = –8 < 0 ∴ The equation has no real roots. x 2 + 2x + 3  0 (x 2 + 2x + 1) – 1 + 3  0 (x + 1) 2 + 2  0 ∵ (x + 1) 2  0 for all real values of x. ∴ (x + 1) 2 + 2 > 0 for all real values of x. ∴ There are no solutions for x 2 + 2x + 3  0.  By completing the square, rewrite x 2 + 2x + 3 into the form (x + h) 2 + k.

25. Follow-up question Solve the quadratic inequality –x 2 + 6x – 10 < 0 by the algebraic method. For the corresponding equation –x 2 + 6x – 10 = 0, = 6 2 – 4(–1)(–10) = –4 < 0 ∴ The equation has no real roots.

26. –x 2 + 6x – 10 < 0 (x 2 – 6x + 3 2 ) – 3 2 + 10 > 0 (x – 3) 2 + 1 > 0 ∵ (x – 3) 2  0 for all real values of x. ∴ (x – 3) 2 + 1 > 0 for all real values of x. ∴ The solutions of –x 2 + 6x – 10 < 0 are all real values of x. x 2 – 6x + 10 > 0  By completing the square, rewrite x 2 – 6x + 10 into the form (x – h) 2 + k. Follow-up question Solve the quadratic inequality –x 2 + 6x – 10 < 0 by the algebraic method.