2. Aims: To be able to solve Modulus equations using graphs and algebraic methods.

3.  Name: Know what I am talking about when I say Modulus Equations/Inequalities  Describe: How to use a GDC to solve modulus equation/inequality problems.  Explain: A shortcut for solving equations algebraically.

4.  Exercises C and D p.31  Prep for next topic: Inverse Trigonometric Functions – see lesson on Core 3 section of mymaths.co.uk  This Lesson’s Mathematician…  Jean-Robert Argand 1768-1862: Swiss bookstore manager and amateur mathematician; he is credited with coining the term module for absolute value (in his many influential works on Complex Numbers) this eventually became known as the “modulus” when written in English.

5.  Take the graphs we have just drawn. y=|x| and y=|2x-4| there are two parts.

7.  b)i) Solutions are where  -(2x-4)=x and  2x-4=x

8.  b)ii) |x| above |2x-4| between the solutions

9.  Match the equations/inequalities with their solutions (make use of a GDC to sketch functions)

10.  Notice that what we end up with…  x=-(2x-4) and x=2x-4 making one side positive and negative leaving the other alone. Obviously the inequalities need a sketch

11.  Solve…  |x-2|=4  x-2=4  x=6  x-2=-4  x=-2 It doesn’t matter which side.

12.  Solve…  |4 - x|=2  4-x=2  x=2  4-x=-2  x=6

13.  Solve…  |x + 1|=|x - 3|  x + 1 = x - 3  0 = -4 (no solution)  x + 1 = 3 - x  x = 1

14.  Solve…  |3x + 1|=|x + 4|  3x + 1 = x + 4  x = 3 / 2  3x + 1 = -x - 4  x = -5 / 4

15.  Solve…  |sinx|= 1 / 2 : -180 ≤ x ≤ 180  sinx = 1 / 2  x = 30, 150  sinx = -1 / 2  x = -30, -150

16.  Solve…  |50 - x 2 |=14  50 - x 2 = 14  x 2 = 36  x = 6, -6  50 - x 2 = -14  x 2 = 64  x= 8, -8