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ACTION RESEARCH ON MATHS...

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- 1. “TRICKS TO REMEMBER THE QUADRATIC EQUATION” ACTION RESEARCH REPORT Guide: Shilpa Popat Investigator: Sugandha Singh School Of science and education Navrachana university Vadodara 2014-2015 B.Ed
- 2. CERTIFICATE This is to certify that Miss Sugandha Singh has completed action research as a part of B.Ed program entitled “Tricks To Remember the Quadratic Rquations”. She has completed it to the best of her capacities and to the best of my Knowledge, this is her original work. Guide: Ms. Shilpa Popat
- 3. Acknowledgement Action research is a process in which participants examine their own educational practice, systematically and carefully, using the techniques of research . Action research approaches students to share their problems, so that teachers will help the students and with activities, shortcut methods, rhymes, songs etc. Help them out from the problem. My sincere thanks go to Mr. P.V. Xavier, principal of Navrachana University who gave me the opportunity to do such a wonderful research and also Ms. Shilpa Popat for her ideas and helped us to think out of the box and make such kind of idea. I am thankful to my co- teacher pupils who motivate me, inspired and helped me while doing my practice teaching and Action Research. Sugandha Singh
- 4. Sr. No. Content Page No. 1 Abstract 2 Introduction 3 Specific Question 4 Research Setting 5 Review 6 Research Method 7 Objective of the study 8 Participants 9 Data Collection Tool 10 Process of Data Collection 11 Analysis of Data 12 Research Finding 13 Conclusion
- 5. Abstract Quadratic Equations An example of a Quadratic Equation: Quadratic Equations make nice curves, like this one: The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x2). It is also called an "Equation of Degree 2" (because of the "2" on the x) Standard Form The Standard Form of a Quadratic Equation looks like this: a, b and c are known values. a can't be 0. "x" is the variable or unknown (we don't know it yet). Here are some more examples: 2x2 + 5x + 3 = 0 In this one a=2, b=5 and c=3
- 6. x2 − 3x = 0 This one is a little more tricky: Where is a? Well a=1, and we don't usually write "1x2" b = -3 And where is c? Well c=0, so is not shown. 5x − 3 = 0 Oops! This one is not a quadratic equation: it is missing x2 (in other words a=0, which means it can't be quadratic) Hidden Quadratic Equations! So the "Standard Form" of a Quadratic Equation is ax2 + bx + c = 0 But sometimes a quadratic equation doesn't look like that! For example: In disguise → In Standard Form a, b and c x2 = 3x -1 Move all terms to left hand side x2 - 3x + 1 = 0 a=1, b=-3, c=1 2(w2 - 2w) = 5 Expand (undo the bracket), and move 5 to left 2w2 - 4w - 5 = 0 a=2, b=-4, c=-5 z(z-1) = 3 Expand, and move 3 to left z2 - z - 3 = 0 a=1, b=-1, c=-3 5 + 1/x - 1/x2 = 0 Multiply by x2 5x2 + x - 1 = 0 a=5, b=1, c=-1
- 7. Have a Play With It Play with the " Quadratic Equation Explorer " so you can see: the graph it makes, and the solutions (called "roots"). How To Solve It? The "solutions" to the Quadratic Equation are where it is equal to zero. There are usually 2 solutions (as shown in the graph above). They are also called "roots", or sometimes "zeros" There are 3 ways to find the solutions: 1. We can Factor the Quadratic (find what to multiply to make the Quadratic Equation) 2. We can Complete the Square, or 3. We can use the special Quadratic Formula: Just plug in the values of a, b and c, and do the calculations. We will look at this method in more detail now.
- 8. About the Quadratic Formula Plus/Minus First of all what is that plus/minus thing that looks like ±? The ± means there are TWO answers: Here is why we can get two answers: But sometimes we don't get two real answers, and the "Discriminant" shows why ... Discriminant Do you see b2 - 4ac in the formula above? It is called the Discriminant, because it can "discriminate" between the possible types of answer: when b2 - 4ac is positive, we get two Real solutions when it is zero we get just ONE real solution (both answers are the same) when it is negative we get two Complex solutions Complex solutions? Let's talk about them after we see how to use the formula. Using the Quadratic Formula Just put the values of a, b and c into the Quadratic Formula, and do the calculations.
- 9. Example: Solve 5x² + 6x + 1 = 0 Coefficients are: a = 5, b = 6, c = 1 Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a Put in a, b and c: x = [ -6 ± √(62-4×5×1) ] / (2×5) Solve: x = [ -6 ± √(36-20) ]/10 x = [ -6 ± √(16) ]/10 x = ( -6 ± 4 )/10 x = -0.2 or -1 Answer: x = -0.2 or x = -1 And we see them on this graph. Check -0.2: 5×(-0.2)² + 6×(-0.2) + 1 = 5×(0.04) + 6×(-0.2) + 1 = 0.2 -1.2 + 1 = 0
- 10. Check -1: 5×(-1)² + 6×(-1) + 1 = 5×(1) + 6×(-1) + 1 = 5 - 6 + 1 = 0 Remembering the Formula I don't know of an easy way to remember the Quadratic Formula, but a kind reader suggested singing it to "Pop Goes the Weasel": ♫ "x equals minus b ♫ "All around the mulberry bush plus or minus the square root The monkey chased the weasel of b-squared minus four a c The monkey thought 'twas all in fun all over two a" Pop! goes the weasel" Try singing it a few times and it will get stuck in your head! Complex Solutions? When the Discriminant (the value b2 - 4ac) is negative we get Complex solutions ... what does that mean? It means our answer will include Imaginary Number . Wow! Example: Solve 5x² + 2x + 1 = 0 Coefficients are: a = 5, b = 2, c = 1 Note that The Discriminant is negative: b2 - 4ac = 22 - 4×5×1 = -16
- 11. Use the Quadratic Formula: x = [ -2 ± √(-16) ] / 10 The square root of -16 is 4i (i is √-1, read Imaginary Numbers to find out more) So: x = ( -2 ± 4i )/10 Answer: x = -0.2 ± 0.4i The graph does not cross the x-axis. That is why we ended up with complex numbers. In some ways it is easier: we don't need more calculation, just leave it as -0.2 ± 0.4i. Summary Quadratic Equation in Standard Form: ax2 + bx + c = 0 Quadratic Equations can be factored Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a When the Discriminant (b2-4ac) is: positive, there are 2 real solutions zero, there is one real solution negative, there are 2 complex solutions
- 12. Learning Quadratic equations (by original method) A quadratic equation is an equation where the highest exponent of a variable is 2. There are three main ways to solve quadratic equations: to factor the quadratic equation if you can do so, to use the quadratic formula, or to complete the square. If you want to know how to master these three methods, just follow these steps steps. Method 1 of 3 : Factoring the Equation 1Combine all of the like terms and move them to one side of the equation. The first step to factoring an equation is to move all of the terms to one side of the equation, keeping the x2 term positive. To combine the terms, add or subtract all of the x2 terms, the x terms, and the constants (integer terms), moving them to one side of the equation so that nothing remains on the other side. Once the other side has no remaining terms, you can just write "0" on that side of the equal sign. Here's how you do it:[1] 2x2 - 8x - 4 = 3x - x2 = 2x2 +x2 - 8x -3x - 4 = 0 3x2 - 11x - 4 = 0 2 Factor the expression. To factor the expression, you have to use the factors of the x2 term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it: Since 3x2 only has one set of possible factors, 3x and x, you can write those in the parenthesis: (3x +/- ? )(x +/- ?) = 0. Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. Try out (3x +1)(x -4). When you multiply them out, you get - 3x2 -12x +x -4. If you combine the terms -12x and x, you get -11x, which is the middle term you were aiming for. You have just factored the quadratic equation.
- 13. As an example, let's try a factoring combination that does not work: (3x -2)(x +2) = 3x2 +6x - 2x -4. If you combine those terms, you get 3x2 -4x -4. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you wanted to get -11x, not -4x. 3 Set each set of parenthesis equal to zero as separate equations. This will lead you to find two values for x that will make the entire equation equal to zero. Now that you've factored the equation, all you have to do is put each set of parenthesis equal to zero. So, you should write 3x +1 = 0 and x - 4 = 0. 4 Solve each equation independently. In a quadratic equation, there will be two given values for x. Just solve each equation independently by isolating the variable and write down the two solutions for x. Here's how you do it: 3x + 1 = 0 = 3x = -1 = 3x/3 = -1/3 x = -1/3 x - 4 = 0 x = 4 x = (-1/3, 4) Method 2 of 3: Using the Quadratic formula 1Combine all of the like terms and move them to one side of the equation.Move all of the terms to one side of the equal sign, keeping the x2 term positive. Write the terms in descending order of degrees, so that the x2 term comes first, followed by the x term and the constant term. Here's how you do it: 4x2 - 5x - 13 = x2 -5 4x2 - x2 - 5x - 13 +5 = 0 3x2 - 5x - 8 = 0
- 14. 2 Write down the quadratic formula. The quadratic formula is: {-b +/-√ (b2 - 4ac)}/2a[2] 3 Identify the values of a, b, and c in the quadratic equation. The variable a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant. For the equation 3x2 -5x - 8 = 0, a = 3, b = -5, and c = -8. Write this down. 4 Substitute the values of a, b, and d into the equation. Now that you know the values of the three variables, you can just plug them into the equation like this: {-b +/-√ (b2 - 4ac)}/2 {-(-5) +/-√ ((-5)2 - 4(3)(-8))}/2(3) = {-(-5) +/-√ ((-5)2 - (-96))}/2(3) 5 Do the math. After you've plugged in the numbers, do the remaining math to simplify positive or negative signs, multiply, or square the remaining terms. Here's how you do it: {-(-5) +/-√ ((-5)2 - (-96))}/2(3) = {5 +/-√(25 + 95)}/6 {5 +/-√(120[[Media:Media:Example.ogg]])}/6 6 Simplify the square root. If the number under the radical symbol is a perfect square, you will get a whole number. If the number is not a perfect square, then simplify to its simplest radical version. If the number is negative, and you're sure it's supposed to be negative, then the roots will be complex. In this example, √(121) = 11. You can write that x = (5 +/- 11)/6. 7 Solve for the positive and negative answers. If you've eliminated the square root symbol, then you can keep going until you've found the positive and negative results for x. Now that you have (5 +/- 11)/6, you can write two options: (5 + 11)/6
- 15. (5 - 11)/6 8 Solve for the positive and negative answers. Just do the math: (5 + 11)/6 = 16/6 (5-11)/6 = -6/6 9 Simplify. To simplify each answer, just divide them by the largest number that is evenly divisible into both numbers. Divide the first fraction by 2, and divide the second by 6, and you have solved for x. 16/6 = 8/3 -6/6 = -1 x = (-1, 8/3) Method 3 of 3: Completing the square 1 Move all of the terms to one side of the equation. Make sure that the a or x2term is positive. Here's how you do it:[3] 2x2 - 9 = 12x = 2x2 - 12x - 9 = 0 In this equation, the a term is 2, the b term is -12, and the c term is -9. 2 Move the c term or constant to the other side. The constant term is the numerical term without a variable. Move it to the right side of the equation: 2x2 - 12x - 9 = 0 2x2 - 12x = 9
- 16. 3 Divide both sides by the coefficient of the a or x2 term. If x2 has no term in front of it, and just has a coefficient of 1, then you can skip this step. In this case, you'll have to divide all of the terms by 2, like so: 2x2/2 - 12x/2 = 9/2 = x2 - 6x = 9/2 4 Divide b by two, square it, and add the result to both sides. The b term in this example is -6. Here's how you do it: -6/2 = -3 = (-3)2 = 9 = x2 - 6x + 9 = 9/2 + 9 5 Simplify both sides. Factor the terms on the left side to get (x-3)(x-3), or (x-3)2. Add the terms on the right side to get 9/2 + 9, or 9/2 + 18/2, which adds up to 27/2. 6 Find the square root of both sides. The square root of (x-3)2 is simply (x-3). You can write the square root of 27/2 as ±√(27/2). Therefore, x - 3 = ±√(27/2). 7 Simplify the radical and solve for x. To simplify ±√(27/2), look for a perfect square within the numbers 27 or 2 or in their factors. The perfect square 9 can be found in 27, because 9 x 3 = 27. To take 9 out of the radical sign, pull out the number 9 from the radical, and write the number 3, its square root, outside the radical sign. Leave 3 in the numerator of the fraction under the radical sign, since that factor of 27 cannot be taken out, and leave 2 on the bottom. Then, move the constant 3 on the left side of the equation to the right, and write down your two solutions for x: x = 3 + (3√6)/2 x = 3 - (3√6)/2
- 17. Tips As you can see,the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead,we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. If the number under the square root is not a perfect square,then the last few steps run a little differently. Here is an example: If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Learning QUADRATIC EQUATIONS (by simple short-cut trick method) You can get good score only by practicing more and more. All you need to do is to do math problems correctly within time, and this can be achieved only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. So Quadratic equations shortcut tricks here for those people. We try our level best to put together all types of shortcut methods here. Example: ax2 + bx + c = 0 it is equal to 0 and the a , b , c are the constant value and we can say that x represent as unknown . The a , b , c are the constant and quadratic coefficient or linear coefficient. Quadratic equation hold the only power of x which is also non negative integer . Example some quadratic equation: Example 1: 6x2 +11x + 3 = 0 Answer : Shortcut tricks : This equation +6 is cofficient of x2 .
- 18. + 11 is cofficient of x +3 is constant term . Step 1: we multiply (+6) x (+3) = +18 Step 2 : we break + 18 in two parts such that addition between them is 11. So , +18 = 9 + 2 = 11 . and product of both factors is 18 . So , +9 and +2 = Sum of is +11 . Step 3:Change the sign of both the factors , So +9 = -9 and +2 = -2 . and divide by cofficient of x2 , So we get -9 / 6 = -3 / 2 and -2 / 6 = – 1 / 3 . Example 2: 4y2 + 12y + 8 = 0 Answer : Shortcut tricks : 4 x 8 = +32 we break + 32 in two parts such that addition between them is 12. +32 = (+8) + (+ 4) = +12 . Change sign of both factor and divide by cofficient of y2 , So – 8 / 4 = – 2 . – 4 / 4 = -1 . i. e . – 2 , – 1 . Example 3: 6 x2 + 10x + 4 = 0 Answer : Shortcut tricks : 6 x 4 = +24 we break + 24 in two parts such that addition between them is 10. +32 = (+6) + (+ 4) = +10 . Change sign of both factor and divide by cofficient of x2 , So , -6 / 6 = -1 . – 4 / 6 = -2 / 3 .
- 19. i. e . -1 , – 2 / 3 . Example 4: x2 + 9x + 20 = 0 Answer : Shortcut tricks : 5 x 4 = +20 we break + 20 in two parts such that addition between them is 9. +20 = (+5) + (+ 4) = +9 . Change sign of both factor and divide by cofficient of x2 , So – 5 / 1 = – 5 – 4 / 1 = – 4. i. e . – 5 , – 4 . Example 5: 3y2 + 19y + 28 = 0 Answer : Shortcut tricks : 3 x 28 = +84 we break + 84 in two parts such that difference between them is 19 +84 = (+12) + (+ 7) = +19 . Change sign of both factor and divide by cofficient of y2 , that is 3 So , y = -12 / 3 = – 4 , y = -7 / 3 = – 7 / 3 Y = – 4 , – 7 / 3
- 20. Research Method Action research is either research initiated to solve the immediate problem or a reflective process of progressive problem solving led by individuals working with others in terms or as part of a “community of practice” to improve the way they address issues and solve problems .Action research involves actively participating in a change situation, often via an existing organization, whilst simultaneously conducting research. Sometimes it is called practitioner based research (McNiff, 2002), is a powerful tool for change and improvement at the local level. Kurt Lewin’s own work (one of action research’s founding fathers) was deliberately intended to change the life chances of disadvantaged groups in terms of housing, employment, prejudices, socialisation and training. Acton research is an interactive inquiry process that balance problem solving actions implemented in a collaborative context with data-driven collaborative analysis or research to understand underlying causes enabling future predictions about personal and organizational change (Reason & Bradbury, 2002) I used the “Problem Solving Method” and “Survey Method” for my action research because it gather data at a particular point in time with the intention of describing the nature of existing conditions, or identifying standard against which existing condition can be compared, or determining the relationship that exist between specific events. Thus, these methods may vary in their level of complexity from those which provide simple frequency counts then those which present relation analysis. The process of working through detailed of a problem to reach a solution. Problem solving may include mathematical or systematic operations and can be a gauge of an individual's critical thinking skills. Some of the causes to use these two methods are- Gather data on a one-short basis and hence is economical and efficient. Generation numerical data. Provide descriptive, inferential and explanatory information. Gathered standardized information i.e. using the same instruments and questions for all participants. lack of clarity of the situation multiple goals
- 21. Objectives of the study (i) To enable the learner to identify, implement, solve and memorize the concept of quadratic equation. (ii) To enable the learner to remember the formula regarding quadratic equation. CONCLUSION Quadratic equation contains formula, rules, problem solving Questions etc. These are the chapter which requires more activity, shortcut tricks, proper implementation of formulas etc. and this is reason that a child does not able to learn this kind of chapter, that a child feels boring and tough to remember. For this kind of topics a teacher can show activity, videos, new methods etc. These all can help the children to take interest in these kinds of topics. When a child feel difficulty to learn a subject, it automatically hates that subject so, a teacher is one who take extra care for this kind topics and help the children’s to understand such kind of topics in a easiest way. I personally feel that this chapter is tough for students if we show them activities, learning formulas in a simple way while connecting them to real life example related, pictures help the children out with this kind of problem. I tried whatever I can do for this age group and help them and make them understood this kind of chapter. Thus, I can say, combine efforts of student interest and teacher’s ideas can solve anything. Innovative and enthusiastic efforts are essential to enhance the performance. Thus we can say that action research, contributes to the theory & knowledge base to enhance practice, supports the professional development of practitioners,builds a collegial networking system,helps practitioners identify problems & seek solutions systematically,can be used at all levels & in all areas of education.
- 22. BIBLIOGRAPHY AND URLs USED Mathematics book (NCERT) http://www.wikihow.com/Factor-Second-Degree-Polynomials-(Quadratic-Equations) http://www.tricki.org/article/How_to_solve_quadratic_equations http://www.mathematicsmagazine.com/corresp/ThomasNguyen/ThomasNguyen- QuadraticEquations.html http://www.math-shortcut-tricks.com/basic-math-shortcut-tricks/