Lecture 2 Slides

# Lecture 2 Slides - Problem .Theareaortheroomis 20m2. Letx...

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9/15/2010 1 Problem: A room is 4m longer than it is wide. The area or the room is 20m 2 . What is the area of the room? Let x be the width of the room. Then the area of the room is x ( x + 4) = x 2 + 4x Equating this to the given area gives x 2 + 4x = 20 Rearranging gives x 2 + 4x–20 = 0 This is a “root finding” problem. Root Finding Problems: General form: find x such that f ( x ) = 0 The values of x for which f ( x ) = 0 are the roots of f ( x ) c bx ax x f ) ( 2 For our problem f ( x ) happens to be a quadratic . The roots can be found using the quadratic formula. a ac b b x x roots 2 4 , 2 2 1 In general there are two roots. One is obtained by using + in the formula and the other by using . If the quantity under the square root is zero the roots are equal. If this quantity is negative the roots are complex numbers.

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9/15/2010 2 Casio Calculator Note Casio calculators can solve quadratics. Hit Mode until menu include “EQN”. Then select this option. Use the right arrow to move from “Unknowns?” to “Degree?” Enter 2 (a quadratic is a second degree polynomial) Enter values for a , b , and c (hit “=“ after each value) Use the up and down arrows to move between the two solutions. If the roots are complex numbers shift plus “=“ toggles between the real and imaginary parts of each solution. If the roots are real this key combination has no effect. The display does not indicate the roots are real or complex (use
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## This note was uploaded on 10/18/2010 for the course ECOR2606 2606 taught by Professor Aaaa during the Spring '10 term at Carleton CA.

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Lecture 2 Slides - Problem .Theareaortheroomis 20m2. Letx...

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