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Published byMaude Webster Modified over 9 years ago
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Aims: To be able to solve Modulus equations using graphs and algebraic methods.
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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.
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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.
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Take the graphs we have just drawn. y=|x| and y=|2x-4| there are two parts.
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b)i) Solutions are where -(2x-4)=x and 2x-4=x
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b)ii) |x| above |2x-4| between the solutions
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Match the equations/inequalities with their solutions (make use of a GDC to sketch functions)
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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
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Solve… |x-2|=4 x-2=4 x=6 x-2=-4 x=-2 It doesn’t matter which side.
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Solve… |4 - x|=2 4-x=2 x=2 4-x=-2 x=6
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Solve… |x + 1|=|x - 3| x + 1 = x - 3 0 = -4 (no solution) x + 1 = 3 - x x = 1
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Solve… |3x + 1|=|x + 4| 3x + 1 = x + 4 x = 3 / 2 3x + 1 = -x - 4 x = -5 / 4
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Solve… |sinx|= 1 / 2 : -180 ≤ x ≤ 180 sinx = 1 / 2 x = 30, 150 sinx = -1 / 2 x = -30, -150
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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
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