Algebra presentation on topic modulus function and polynomials
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Algebra Presentation on Topic Modulus Function and Polynomials Created by : Jocelyn AS and Alayya MH Performance Task Binus School Bekasi
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Nature of the roots and sum and product of the roots of a quadratic equation
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This document provides an overview of quadratic equations and inequalities. It defines quadratic equations as equations of the form ax2 + bx + c = 0, where a, b, and c are real number constants and a ≠ 0. Examples of quadratic equations are provided. Methods for solving quadratic equations are discussed, including factoring, completing the square, and the quadratic formula. Properties of inequalities are outlined. The chapter also covers solving polynomial and rational inequalities, as well as equations and inequalities involving absolute value. Practice problems are included at the end.
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This document discusses solving quadratic inequalities. It provides examples of single-variable quadratic inequalities and explains how to find the solution set by first setting the inequality equal to an equation, solving for the roots, and then testing values within the intervals formed by the roots. The document also introduces quadratic inequalities with two variables and how to represent them. It defines a quadratic inequality and explains the process of solving them by relating it back to solving a quadratic equation.
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1. A test on polynomials will be held on Thursday covering adding, subtracting, multiplying, and dividing polynomials.
2. Test scores from additional tests will be posted with the lowest score being dropped. Weekly Khan Academy grades for the first two weeks of the third quarter will be posted after Wednesday.
3. The new scoring schedule for Khan Academy assignments is 4 points for each correct answer, with 8 consecutive correct answers earning 100% and 7 out of 8 correct earning 28%. No credit will be given for questions where hints are used.
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- 1. By Alayya M.H and Jocelyn A.S Young’s Modulus AND POLYNOMIALS
- 2. Polynomials Question The polynomial 4x3+ax2-bx-2, is denoted by p(x). The results of differentiating p(x) with respect to x is denoted by p’(x) and the second derivative is p’’(x). It is given that when p’(x) is divided by (x+2) the remainder is -1 and when p”(x) is divided by (x+1) the remainder is -2. (i) Find the values of a and b (ii) When a and b have these values, find the remainder when p(x) is divided by (x+1)
- 3. How To Solve (i)? y= 4x3+ax2-bx-2 dy/dx = 12x2+2ax-b p’(x)= 12x2+2ax-b Step 1: Find the first derivative Step 2: Input the value of x into the First derivative x+2=0 X = -2 p’(x) = 12x2+2ax-b p’(-2) = 12(-2)2 + 2a(-2) -b -1 = 48 -4a-b 49 =4a +b
- 4. dy/dx = 12x2+2ax-b d2y/dx2 = 24x + 2a p’’(x)= 24x + 2a Step 3: Find the second derivative Step 4: Input the value of x into the second derivative X+1 = 0 X = -1 p’’(x)= 24x + 2a p’’(-1)= 24(-1) + 2a -2 = -24 +2a 22 = 2a 11 = a
- 5. Step 5: Substitute the value of A into the equation found in step 2 to find b A = 11 49 = 4a +b 49 = 4(11) + b 49 = 44 + b 5 = b Thus, A is 11 and B is 5
- 6. Since a is 11 and b is 5, p(x) = 4x3+11x2-5x-2 How To Solve (ii)? 4x3+11x2-5x-2X +1 4x3+4x2 7x2-5x -2 7x2+7x - 12x-2 -12x-12 10 Remainder 4x2+7x-12
- 7. Young’s modulus question 2 Questions: 1. Solve the inequality |3x - 4| < |3x - 6| 2. Hence find the largest integer n satisfying the inequality |3 ln n-4| < |3 ln n- 6 |
- 8. How to solve number 1? Step 1: square both sides of the equation (3x - 4)2 < (3 x - 6)2 Step 2: expand both sides of the equation 9x2 - 24x + 16 < 9x2 -36x + 36
- 9. Step 3 Simplify the equation and find the x ! 9x2- 9x2 -24 x + 36 x + 16-36 <0 12 x - 20 < 0 12 x < 20 X < 5/3
- 10. How to solve number 2 ? Step 1 : turn the solution into “ln n “ Step 2 : Find the largest integer Let ln n = X From the solution we get X < 5/3. So ln n < 5/3 ln n = 5/3 ln n = logen logen= 5/3 n = e ^ 5/3 e = natural number n= 5.294
- 11. Step 3: Round=off As we know from the initial equation in slide 7 that it is less than, then we need to round it off to 5 (from 5.294). Thus, n=5 . Do not round off to 6 because this is less than not more than.
- 12. 1. Template : Slidesgo.com 2. Problem inspiration (for polynomials): (9709-W 2009-Paper 3/2-Q5) References
- 13. Thank you