DOCSLIB.ORG

SOLVING QUADRATIC EQUATIONS Factoring Method Square Root Property

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
SOLVING QUADRATIC EQUATIONS Factoring Method Square Root Property SOLVING QUADRATIC EQUATIONS A quadratic equation in is an equation that may be written in the standard quadratic form ͦ if . There are four different methods used to ͬ ͕ͬ ƍ ͖ͬ ƍ ͗ Ɣ 0 ͕ ƕ 0 solve equations of this type. Factoring Method If the quadratic polynomial can be factored, the Zero Product Property may be used. This property states that when the product of two factors equals zero, then at least one of the factors is zero. If and are algebraic expressions, then if and only if or . ̻ ̼ ̻̼ Ɣ 0 ̻ Ɣ 0 ̼ Ɣ 0 Steps to solve quadratic equations by factoring: 1. Write the equation in standard form (equal to 0). 2. Factor the polynomial. 3. Use the Zero Product Property to set each factor equal to zero. 4. Solve each resulting linear equation. Examples: A. B. C. ͬͦ ƍ 5ͬ Ɣ 24 9ͬͦ ƍ 12 Ɣ 3 ƍ 12ͬ ƍ 5ͬͦ 3ͬͦ Ɣ 4 Ǝ 11ͬ 1. 1. 1. ͬͦ ƍ 5ͬ Ǝ 24 Ɣ 0 4ͬͦ Ǝ 12ͬ ƍ 9 Ɣ 0 3ͬͦ ƍ 11ͬ Ǝ 4 Ɣ 0 2. 2. 2. ʚͬ ƍ 8ʛʚͬ Ǝ 3ʛ Ɣ 0 ʚ2ͬ Ǝ 3ʛʚ2ͬ Ǝ 3ʛ Ɣ 0 ʚ3ͬ Ǝ 1ʛʚͬ ƍ 4ʛ Ɣ 0 3. or 3. 3. or ͬ ƍ 8 Ɣ 0 ͬ Ǝ 3 Ɣ 0 2ͬ Ǝ 3 Ɣ 0 3ͬ Ǝ 1 Ɣ 0 ͬ ƍ 4 Ɣ 0 4. or ͧ ͥ ͬ Ɣ Ǝ8 ͬ Ɣ 3 4. 4. or ͬ Ɣ ͦ ͬ Ɣ ͧ ͬ Ɣ Ǝ4 Square Root Property This property states: If and are algebraic expressions such that ͦ , then . This method is used if the form of the equation is: ̻ ̼ ̻ Ɣ ̼ ̻ Ɣ Ə√̼ or (where represents a constant) ͦ ʚ ʛͦ ͬ Ɣ ͟ ͕ͬ ƍ ͖ Ɣ ͟ ͟ Steps to solve quadratic equations by the square root property: 1. Transform the equation so that a perfect square is on one side and a constant is on the other side of the equation. 2. Use the square root property to find the square root of each side. REMEMBER that finding the square root of a constant yields positive and negative values. 3. Solve each resulting equation. (If you are finding the square root of a negative number, there is no real solution and imaginary numbers are necessary.) Examples: C. B. A. ͦ ͬͦ Ǝ 16 Ɣ 0 ʚͬ ƍ 1ʛͦ Ɣ 49 ʚͬ Ǝ 3ʛ Ɣ Ǝ28 1. ͦ 1. ͦ 1. ͦ ͬ Ɣ 16 ʚͬ ƍ 1ʛ Ɣ 49 ʚͬ Ǝ 3ʛ Ɣ Ǝ28 2. ͦ 2. 2. ͦ √ ͬ ƔƏ√16 ǭʚͬ ƍ 1ʛͦ ƔƏ√49 ǭʚͬ Ǝ 3ʛ ƔƏ√Ǝ28 3. 3. 3. ͬ Ɣ Ə4 ͬ ƍ 1 Ɣ 7 or ͬ ƍ 1 Ɣ Ǝ7 ͬ Ǝ 3 Ɣ Ə2͝√7 ͬ Ɣ 6 or ͬ Ɣ Ǝ8 ͬ Ɣ 3 Ə 2͝√7 Completing the Square This method may be used to solve all quadratic equations. ( ͦ ) ͕ͬ ƍ͖ͬƍ͗Ɣ0, ͕ƕ0 Steps to solve an equation by completing the square: 1. Transform the equation so that the quadratic term and the linear term equal a constant. ͕ͬͦ ƍ ͖ͬ Ɣ Ǝ͗ 2. Divide each term by the coefficient of the quadratic term if it is not a one. ͬͦ ƍ ͬ Ɣ Ǝ 3. Complete the square: • Multiply the coefficient of by ͥ or divide it by 2. ( ) ͬ ͦ ͦ ͦ • Square this value. ʠͦʡ ͦ ͦ • Add the result to both sides of the equation. ͦ ͬ ƍ ͬ ƍ ʠͦʡ ƔƎ ƍ ʠͦʡ • Express one side of the equation as the square of a binomial and the other as a constant. ͦ where is the constant formed by ͦ. ʠͬ ƍ ʡ Ɣ ͟ ͟ Ǝ ƍ ʠ ʡ 4. Use the square root property to find the square root of each side of the equation. ͦ ͦ 5. Solve each resulting equation. Examples: A. B. C. ͬͦ ƍ 12ͬ ƍ 2 Ɣ 0 ͬͦ Ɣ 2ͬ Ǝ 6 2ͬͦ ƍ 8ͬ Ǝ 15 Ɣ 0 1. 1. 1. ͬͦ ƍ 12ͬ Ɣ Ǝ2 ͬͦ Ǝ 2ͬ Ɣ Ǝ6 2ͬͦ ƍ 8ͬ Ɣ 15 2. Lead coefficient is a one. 2. Lead coefficient is a one. 2. ͦ ͥͩ ͬ ƍ 4ͬ Ɣ ͦ ͦ ͦ ͦ 3. 12 3. Ǝ2 ͥͩ 4 ͦ ƴ Ƹ Ɣ 36 ͦ ƴ Ƹ Ɣ 1 3. ͦ ƴ Ƹ Ɣ 4 ͬ ƍ 12ͬ ƍ 36 Ɣ Ǝ2 ƍ 36 2 ͬ Ǝ 2ͬ ƍ 1 Ɣ Ǝ6 ƍ 1 2 ͬ ƍ 4ͬ ƍ 4 Ɣ ͦ ƍ 4 2 ͦ ͦ ͦ ͦͧ ʚͬ ƍ 6ʛ Ɣ 34 ʚͬ Ǝ 1ʛ Ɣ Ǝ5 ʚͬ ƍ 2ʛ Ɣ ͦ 4. ͦ 4. ͦ 4. ͦͧ ǭʚͬ ƍ 6ʛ ƔƏ√34 ǭʚͬ Ǝ 1ʛ ƔƏ√Ǝ5 ͦ ǭʚͬ ƍ 2ʛ Ɣ Əǯ ͦ ͬ ƍ 6 Ɣ Ə√34 ͬ Ǝ 1 Ɣ Ə͝√5 ͦͧ ͬ ƍ 2 Ɣ Əǯ ͦ 5. 5. ͬ Ɣ Ǝ6 Ə √34 ͬ Ɣ 1 Ə ͝√5 5. √ͨͪ ͬ Ɣ Ǝ2 Ə ͦ Quadratic Formula The quadratic formula, which also may be used to solve any quadratic equation, results from solving the quadratic equation for by ͕ͬͦ ƍ͖ͬƍ͗Ɣ0, ͕ƕ0 ͬ completing the square. ͦ Ǝ͖ Ə √͖ Ǝ 4͕͗ ͬ Ɣ 2͕ Examples: A. B. ͬͦ ƍ 6ͬ Ɣ 16 Ǝͬͦ Ǝ 4ͬ ƍ 8 Ɣ 0 ͬͦ ƍ 6ͬ Ǝ 16 Ɣ 0 ͬͦ ƍ 4ͬ Ǝ 8 Ɣ 0 ͦ ͦ Ǝ͖ Ə √͖ Ǝ 4͕͗ Ǝ͖ Ə √͖ Ǝ 4͕͗ ͬ Ɣ ͬ Ɣ 2͕ 2͕ ͦ ͦ Ǝ6 Ə ǭ6 Ǝ 4ʚ1ʛʚƎ16 ʛ Ǝ4 Ə ǭ4 Ǝ 4ʚ1ʛʚƎ8ʛ ͬ Ɣ ͬ Ɣ 2ʚ1ʛ 2ʚ1ʛ Ǝ6 Ə √100 Ǝ4 Ə √48 ͬ Ɣ ͬ Ɣ 2 2 Ǝ6 Ə 10 Ǝ4 Ə 4√3 ͬ Ɣ ͬ Ɣ 2 2 ͬ Ɣ 2 or ͬ Ɣ Ǝ8 ͬ Ɣ Ǝ2 Ə 2√3 Practice Problems 1. 15. 29. ͬͦ ƍ 6ͬ ƍ 9 Ɣ 0 ͬͦ Ǝ 2ͬ ƍ 1 Ɣ 0 6ʚͬ ƍ 1ʛͦ ƍ 7ͬ Ɣ Ǝ9 2. 16. 30. 5ͬͦ Ɣ 13ͬ ƍ 6 2ͬͦ Ǝ 8ͬ ƍ 8 Ɣ 0 ʚͬ Ǝ 1ʛʚͬ ƍ 1ʛ Ɣ 5ʚͬ ƍ 1ʛ 3. 17. 31. ͬͦ ƍ 25 Ɣ 0 Ǝ9ͭͦ Ǝ 6ͭ Ǝ 1 Ɣ 0 ͬͦ ƍ 25 Ɣ 10ͬ 4. 18. ͭ ͬͦ Ǝ ͬ ƍ 4 Ɣ 0 4͕ͦ Ǝ 12͕ ƍ 9 Ɣ 0 32. 5. 19. ͭ Ɣ ͪͯ4 ͬͦ Ǝ 2ͬ Ǝ 8 Ɣ 0 ͬͦ Ǝ 2ͬ Ǝ 4 Ɣ 0 33. 6. ͦ 20. ͦ 3ͦʚͦ Ǝ 5ʛ ƍ 16 Ɣ Ǝͦ ͬ Ǝ 2ͬ ƍ 2 Ɣ 0 ͕ ƍ ͕ Ǝ 5 Ɣ 0 34. ͦ 7. ͦ 21. ͦ ͬ ƍ 4ʚͬ ƍ 1ʛ Ɣ 0 2ͬ Ǝ ͬ Ǝ 1 Ɣ 0 Ǝͦ ƍ ͦ ƍ 1 Ɣ 0 35. ͦ 8. ͦ 22. ͦ ͬ Ǝ 3 Ɣ ͬ ͬ Ɣ 4ͬ Ǝ 12 ͬ Ǝ 2 Ɣ 0 36. ͧ 9. ͦ 23. ͦ ͬ ƍ ͬ Ǝ 4 Ɣ 0 ͬ ƍ ͬ ƍ 1 Ɣ 0 ͬ ƍ 3 Ɣ 3 10. 24. 37. ͬͦ ƍ 10 Ɣ 6ͬ Ǝͮͦ ƍ 2ͮ Ǝ 7 Ɣ 0 3͕ʚ͕ ƍ 2ʛ ƍ 1 Ɣ 0 11. ͦ 25. ͦ 38. ͥͯͩ( ͬ Ǝ 7ͬ ƍ 6 Ɣ 0 ͕ ƍ 1 Ɣ 0 ͦ͡ ƍ 1 Ɣ 12. ͦ 26. ͦ ͦ Ǝͬ ƍ ͬ ƍ 6 Ɣ 0 4ͬ ƍ 4ͬ ƍ 5 Ɣ 0 39. ͯͨ 13. ͦ 27. ͦ ͬ Ɣ 3 Ǝ3͕ ƍ 5͕ Ǝ 2 Ɣ 0 3ͬ Ɣ 4ͬ ƍ 7 14. ͦ ͩ 3ͮͧ 40. 5ͭ Ǝ ͭ Ɣ 0 28. ʚͬ ƍ 9ʛʚͬ Ǝ 3ʛ Ɣ Ǝ37 ͬ ƍ 3 Ǝ 3 Ɣ 3 Practice Problems Answers 1. ͬ Ɣ Ǝ3 23. ͥ √ͧ ͬ Ɣ Ǝ ͦ Ə ͦ ͝ 2. ͦ ͬ Ɣ 3 or ͬ Ɣ Ǝ 24. ͩ ͮ Ɣ 1 Ə ͝√6 3. 25. ͬ Ɣ Ə5͝ ͕ Ɣ Ə͝ 4. ͥ √ͥͩ ͥ ͬ Ɣ Ə ͝ 26. ͦ ͦ ͬ Ɣ Ǝ ͦ Ə ͝ 5. 27. ͫ ͬ Ɣ 4 or ͬ Ɣ Ǝ2 ͬ Ɣ Ǝ1 or ͬ Ɣ 6. ͧ 1 Ə ͝ 28. ͬ Ɣ Ǝ4 or ͬ Ɣ 2 7. ͥ ͬ Ɣ 1 or ͬ Ɣ Ǝ 29. ͧ ͩ ͦ ͬ Ɣ Ǝ or ͬ Ɣ Ǝ 8. ͦ ͧ ͬ Ɣ 2 Ə 2͝ 2 30. √ ͬ Ɣ 6 or ͬ Ɣ Ǝ1 9. ͯͥƏ√ͥͫ 31. ͬ Ɣ ͦ ͬ Ɣ 5 10. 32. ͬ Ɣ 3 Ə ͝ ͭ Ɣ 3 11. 33. ͬ ͬ Ɣ 6 or ͬ Ɣ 1 ͦ Ɣ ͧ or ͦ Ɣ 2 12. 34. ͬ Ɣ 3 or ͬ Ɣ Ǝ2 ͬ Ɣ Ǝ2 13. ͦ 35. ͥƏ√ͥͧ ͕ Ɣ 1 or ͬ Ɣ ͧ ͬ Ɣ ͦ 14. ͥ ͭ Ɣ 0 or ͭ Ɣ 36. ͯͧƏ√ͦͥ ͩ ͬ Ɣ 15. ͦ ͬ Ɣ 1 37. ͯͧƏ√ͪ 16. ͕ Ɣ ͬ Ɣ 2 ͧ 17. ͥ 38. ͯͩƏ√ͥͫ ͭ Ɣ Ǝ ͧ ͡ Ɣ ͨ 18. ͧ 39. ͕ Ɣ ͬ Ɣ Ə2͝ ͦ 40. 19. ͬ Ɣ Ǝ3 Ə ͝ ͬ Ɣ 1 Ə √5 20. ͯͥƏ√ͦͥ ͕ Ɣ ͦ 21. ͥƏ√ͩ ͦ Ɣ ͦ 22. ͬ Ɣ Ə√2.
Load more

Recommended publications
  • 6 Linear and Quadratic Functions
    6 Linear and Quadratic Functions 6.1 Prologue Definition 4 in Book I of Euclid’s Elements (3rd century BC) reads: “A straight line is a line that lies evenly with the points on itself.” No one knows for sure what exactly he meant by this — maybe the meaning got lost in translation. Perhaps he meant that if you aim from one point to another, all the points in between fall on that line and no points stick out. In any case, we know what a straight line is. Or do we? A stretched piece of string? A ray of light? This is probably the most deceptive intuitive notion. Einstein’s theory of relativity states that rays of light are actually bent by gravity. In general, our faith in Euclidean geometry has been shaken by more recent (18th century) models of non-Euclidean geometries, in which there exists more than one line through a given point that is parallel to a given line, or in which no parallel lines exist at all. What if our space is actually curved? Here we’ll avoid these mind-bending considerations and stick with the Euclidean point of view. We define a straight line on the Cartesian plane as a graph of a linear relation px+= qy C . If q ≠ 0 , this relation is a function of the form y =+mx b , where p C m =− and b = . A function y = mx+ b is called a linear function. In q q Sections <...>-<...>, we will discuss the properties of linear relations and functions. A function given by the formula f ()xaxbxc= 2 ++, where a, b, and c are constants and a ≠ 0 , is called a quadratic function.
    [Show full text]
  • The Diamond Method of Factoring a Quadratic Equation
    The Diamond Method of Factoring a Quadratic Equation Important: Remember that the first step in any factoring is to look at each term and factor out the greatest common factor. For example: 3x2 + 6x + 12 = 3(x2 + 2x + 4) AND 5x2 + 10x = 5x(x + 2) If the leading coefficient is negative, always factor out the negative. For example: -2x2 - x + 1 = -1(2x2 + x - 1) = -(2x2 + x - 1) Using the Diamond Method: Example 1 2 Factor 2x + 11x + 15 using the Diamond Method. +30 Step 1: Multiply the coefficient of the x2 term (+2) and the constant (+15) and place this product (+30) in the top quarter of a large “X.” Step 2: Place the coefficient of the middle term in the bottom quarter of the +11 “X.” (+11) Step 3: List all factors of the number in the top quarter of the “X.” +30 (+1)(+30) (-1)(-30) (+2)(+15) (-2)(-15) (+3)(+10) (-3)(-10) (+5)(+6) (-5)(-6) +30 Step 4: Identify the two factors whose sum gives the number in the bottom quarter of the “x.” (5 ∙ 6 = 30 and 5 + 6 = 11) and place these factors in +5 +6 the left and right quarters of the “X” (order is not important). +11 Step 5: Break the middle term of the original trinomial into the sum of two terms formed using the right and left quarters of the “X.” That is, write the first term of the original equation, 2x2 , then write 11x as + 5x + 6x (the num bers from the “X”), and finally write the last term of the original equation, +15 , to get the following 4-term polynomial: 2x2 + 11x + 15 = 2x2 + 5x + 6x + 15 Step 6: Factor by Grouping: Group the first two terms together and the last two terms together.
    [Show full text]
  • Solving Cubic Polynomials
    Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial monic. a 2. Then, given x2 + a x + a , substitute x = y − 1 to obtain an equation without the linear term. 1 0 2 (This is the \depressed" equation.) 3. Solve then for y as a square root. (Remember to use both signs of the square root.) a 4. Once this is done, recover x using the fact that x = y − 1 . 2 For example, let's solve 2x2 + 7x − 15 = 0: First, we divide both sides by 2 to create an equation with leading term equal to one: 7 15 x2 + x − = 0: 2 2 a 7 Then replace x by x = y − 1 = y − to obtain: 2 4 169 y2 = 16 Solve for y: 13 13 y = or − 4 4 Then, solving back for x, we have 3 x = or − 5: 2 This method is equivalent to \completing the square" and is the steps taken in developing the much- memorized quadratic formula. For example, if the original equation is our \high school quadratic" ax2 + bx + c = 0 then the first step creates the equation b c x2 + x + = 0: a a b We then write x = y − and obtain, after simplifying, 2a b2 − 4ac y2 − = 0 4a2 so that p b2 − 4ac y = ± 2a and so p b b2 − 4ac x = − ± : 2a 2a 1 The solutions to this quadratic depend heavily on the value of b2 − 4ac.
    [Show full text]
  • Factoring Polynomials
    EAP/GWL Rev. 1/2011 Page 1 of 5 Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. Example: — Set the factors of a polynomial equation (as opposed to an expression) equal to zero in order to solve for a variable: Example: To solve ,; , The flowchart below illustrates a sequence of steps for factoring polynomials. First, always factor out the Greatest Common Factor (GCF), if one exists. Is the equation a Binomial or a Trinomial? 1 Prime polynomials cannot be factored Yes No using integers alone. The Sum of Squares and the Four or more quadratic factors Special Cases? terms of the Sum and Difference of Binomial Trinomial Squares are (two terms) (three terms) Factor by Grouping: always Prime. 1. Group the terms with common factors and factor 1. Difference of Squares: out the GCF from each Perfe ct Square grouping. 1 , 3 Trinomial: 2. Sum of Squares: 1. 2. Continue factoring—by looking for Special Cases, 1 , 2 2. 3. Difference of Cubes: Grouping, etc.—until the 3 equation is in simplest form FYI: A Sum of Squares can 1 , 2 (or all factors are Prime). 4. Sum of Cubes: be factored using imaginary numbers if you rewrite it as a Difference of Squares: — 2 Use S.O.A.P to No Special √1 √1 Cases remember the signs for the factors of the 4 Completing the Square and the Quadratic Formula Sum and Difference Choose: of Cubes: are primarily methods for solving equations rather 1. Factor by Grouping than simply factoring expressions.
    [Show full text]
  • Math 135 Circles and Completing the Square Examples a Perfect Square
    Math 135 Circles and Completing the Square Examples A perfect square is a number a such that a = b2 for some real number b. Some examples of perfect squares are 4 = 22; 16 = 42; 169 = 132. We wish to have a method for finding b when a is an expression. For instance, you should remember that a2 + 2ab + b2 is a perfect square, because it is exactly (a + b)2. How would you turn the expression x2 + ax into a perfect square? a 2 2 A moment of thought should convince you that if we add ( 2 ) to x + ax we obtain a a 2 2 a 2 a 2 perfect square, because (x + 2 ) = x + ax + ( 2 ) . The addition of ( 2 ) is called completing the square, because the new expression can now be written as a square of some other expression. Example 1. Complete the square: x2 + 4x = 0 x2 + 4x = 0 () (x2 + 4x + 4) = 4 () (x + 2)2 = 4 We have added the square of half the coefficient of x to the original equation, and therefore to maintain equality it was necessary to add the same amount to the other side of the equation. Warning 2. The coefficient of x2 must be equal to 1 in order to complete the square. Example 3. Complete the square: 2x2 + 8x = 0 2x2 + 8x = 0 () 2(x2 + 4x) = 0 () 2(x2 + 4x + 4) = 8 () 2(x + 2)2 = 8 We added 4, the square of half the coefficient of x, inside the parentheses. Note that this amounts to adding 8 to the left side of the equation, because everything inside the parentheses is multiplied by 2.
    [Show full text]
  • Algebra II Level 1
    Algebra II 5 credits – Level I Grades: 10-11, Level: I Prerequisite: Minimum grade of 70 in Geometry Level I (or a minimum grade of 90 in Geometry Topics) Units of study include: equations and inequalities, linear equations and functions, systems of linear equations and inequalities, matrices and determinants, quadratic functions, polynomials and polynomial functions, powers, roots, and radicals, rational equations and functions, sequence and series, and probability and statistics. PROFICIENCIES INEQUALITIES -Solve inequalities -Solve combined inequalities -Use inequality models to solve problems -Solve absolute values in open sentences -Solve absolute value sentences graphically LINEAR EQUATION FUNCTIONS -Solve open sentences in two variables -Graph linear equations in two variables -Find the slope of the line -Write the equation of a line -Solve systems of linear equations in two variables -Apply systems of equations to solve real word problems -Solve linear inequalities in two variables -Find values of functions and graphs -Find equations of linear functions and apply properties of linear functions -Graph relations and determine when relations are functions PRODUCTS AND FACTORS OF POLYNOMIALS -Simplify, add and subtract polynomials -Use laws of exponents to multiply a polynomial by a monomial -Calculate the product of two or more polynomials -Demonstrate factoring -Solve polynomial equations and inequalities -Apply polynomial equations to solve real world problems RATIONAL EQUATIONS -Simplify quotients using laws of exponents -Simplify
    [Show full text]
  • Math: Honors Geometry UNIT/Weeks (Not Timeline/Topics Essential Questions Consecutive)
    Math: Honors Geometry UNIT/Weeks (not Timeline/Topics Essential Questions consecutive) Reasoning and Proof How can you make a Patterns and Inductive Reasoning conjecture and prove that it is Conditional Statements 2 true? Biconditionals Deductive Reasoning Reasoning in Algebra and Geometry Proving Angles Congruent Congruent Triangles How do you identify Congruent Figures corresponding parts of Triangle Congruence by SSS and SAS congruent triangles? Triangle Congruence by ASA and AAS How do you show that two 3 Using Corresponding Parts of Congruent triangles are congruent? Triangles How can you tell whether a Isosceles and Equilateral Triangles triangle is isosceles or Congruence in Right Triangles equilateral? Congruence in Overlapping Triangles Relationships Within Triangles Mid segments of Triangles How do you use coordinate Perpendicular and Angle Bisectors geometry to find relationships Bisectors in Triangles within triangles? 3 Medians and Altitudes How do you solve problems Indirect Proof that involve measurements of Inequalities in One Triangle triangles? Inequalities in Two Triangles How do you write indirect proofs? Polygons and Quadrilaterals How can you find the sum of The Polygon Angle-Sum Theorems the measures of polygon Properties of Parallelograms angles? Proving that a Quadrilateral is a Parallelogram How can you classify 3.2 Properties of Rhombuses, Rectangles and quadrilaterals? Squares How can you use coordinate Conditions for Rhombuses, Rectangles and geometry to prove general Squares
    [Show full text]
  • Further Techniques and App Icat Ions
    Chapter 10 Further Techniques and App icat ions Some simple geometric problems require advanced methods of integration. Besides the basic methods of integration associated with reversing the differen- tiation rules, there are special methods for integrands of particular forms. Using these methods, we can solve some interesting length and area problems. 10.1 Trigonometric Integrals The key to evaluating many integrals is a trigonometric identity or substitution. The integrals treated in this section fall into two groups. First, there are some purely trigonometric integrals that can be evaluated using trigonometric identities. Second, there are integrals involving quadratic functions and their square roots which can be evaluated using trigonometric substitutions. We begin by considering integrals of the form J'sinmx cosnxdx, where m and n are integers. The case n = 1 is easy, for if we let u = sinx, we find I sinm+ ' (x) J'sinmxcosxdx= umdu=- +C= +C J' m+l m+l (or lnlsinxl + C, if m = - 1). The case m = 1 is similar: c0sn+ l(x) Jsin x cosnxdx = - +C n+l (or - lnlcosxl + C, if n = - 1). If either m or n is odd, we can use the identity sin2x + cos2x = 1 to reduce the integral to one of the types just treated. Example 1 Evaluate Jsin2x cos3x dx. Solution Jsin2x cos3x dx = Jsin2x cos2xcosx dx = J(sin2x)(l - sin2x)cosx dx, which can be integrated by the substitution u = sinx. We get Copyright 1985 Springer-Verlag. All rights reserved. 458 Chapter 10 Further Techniques and Applications of Integration If m = 2k and n = 21 are both even, we can use the half-angle formulas sin2x = (1 - cos2x)/2 and cos2x = (1 + cos2x)/2 to write where y = 2x.
    [Show full text]
  • Quadratic Equations Through History
    Cal McKeever & Robert Bettinger COMMON CORE STANDARD 3B Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. Factor a quadratic expression to reveal the zeros of the function it defines. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. SYNOPSIS At a summit of time-traveling historical mathematicians, attendees from Ancient Babylon pose the question of how to solve quadratic equations. Their method of completing the square serves their purposes, but there was far more to be done with this complex equation. Pythagoras, Euclid, Brahmagupta, Bhaskara II, Al-Khwarizmi and Descartes join the discussion, each pointing out their contributions to the contemporary understanding of solving quadratics through the quadratic formula. Through their discussion, the historical situation through which we understand the solving of quadratic equations is highlighted, showing the complex history of this formula taught everywhere. As the target audience for this project is high school students, the different characters of the script use language and symbolism that is anachronistic but will help the the students to understand the concepts that are discussed. For example, Diophantus did not use a,b,c in his actual texts but his fictional character in the the given script does explain his method of solving quadratics using contemporary notation. HISTORICAL BACKGROUND The problem of solving quadratic equations dates back to Babylonia in the 2nd Millennium BC. The Babylonian understanding of quadratics was used geometrically, to solve questions of area with real-world solutions.
    [Show full text]
  • The Computational Complexity of Polynomial Factorization
    Open Questions From AIM Workshop: The Computational Complexity of Polynomial Factorization 1 Dense Polynomials in Fq[x] 1.1 Open Questions 1. Is there a deterministic algorithm that is polynomial time in n and log(q)? State of the art algorithm should be seen on May 16th as a talk. 2. How quickly can we factor x2−a without hypotheses (Qi Cheng's question) 1 State of the art algorithm Burgess O(p 2e ) 1.2 Open Question What is the exponent of the probabilistic complexity (for q = 2)? 1.3 Open Question Given a set of polynomials with very low degree (2 or 3), decide if all of them factor into linear factors. Is it faster to do this by factoring their product or each one individually? (Tanja Lange's question) 2 Sparse Polynomials in Fq[x] 2.1 Open Question α β Decide whether a trinomial x + ax + b with 0 < β < α ≤ q − 1, a; b 2 Fq has a root in Fq, in polynomial time (in log(q)). (Erich Kaltofen's Question) 2.2 Open Questions 1. Finding better certificates for irreducibility (a) Are there certificates for irreducibility that one can verify faster than the existing irreducibility tests? (Victor Miller's Question) (b) Can you exhibit families of sparse polynomials for which you can find shorter certificates? 2. Find a polynomial-time algorithm in log(q) , where q = pn, that solves n−1 pi+pj n−1 pi Pi;j=0 ai;j x +Pi=0 bix +c in Fq (or shows no solutions) and works 1 for poly of the inputs, and never lies.
    [Show full text]
  • Section P.2 Solving Equations Important Vocabulary
    Section P.2 Solving Equations 5 Course Number Section P.2 Solving Equations Instructor Objective: In this lesson you learned how to solve linear and nonlinear equations. Date Important Vocabulary Define each term or concept. Equation A statement, usually involving x, that two algebraic expressions are equal. Extraneous solution A solution that does not satisfy the original equation. Quadratic equation An equation in x that can be written in the general form 2 ax + bx + c = 0 where a, b, and c are real numbers with a ¹ 0. I. Equations and Solutions of Equations (Page 12) What you should learn How to identify different To solve an equation in x means to . find all the values of x types of equations for which the solution is true. The values of x for which the equation is true are called its solutions . An identity is . an equation that is true for every real number in the domain of the variable. A conditional equation is . an equation that is true for just some (or even none) of the real numbers in the domain of the variable. II. Linear Equations in One Variable (Pages 12-14) What you should learn A linear equation in one variable x is an equation that can be How to solve linear equations in one variable written in the standard form ax + b = 0 , where a and b and equations that lead to are real numbers with a ¹ 0 . linear equations A linear equation has exactly one solution(s). To solve a conditional equation in x, . isolate x on one side of the equation by a sequence of equivalent, and usually simpler, equations, each having the same solution(s) as the original equation.
    [Show full text]
  • Quadratic Polynomials
    Quadratic Polynomials If a>0thenthegraphofax2 is obtained by starting with the graph of x2, and then stretching or shrinking vertically by a. If a<0thenthegraphofax2 is obtained by starting with the graph of x2, then flipping it over the x-axis, and then stretching or shrinking vertically by the positive number a. When a>0wesaythatthegraphof− ax2 “opens up”. When a<0wesay that the graph of ax2 “opens down”. I Cit i-a x-ax~S ~12 *************‘s-aXiS —10.? 148 2 If a, c, d and a = 0, then the graph of a(x + c) 2 + d is obtained by If a, c, d R and a = 0, then the graph of a(x + c)2 + d is obtained by 2 R 6 2 shiftingIf a, c, the d ⇥ graphR and ofaax=⇤ 2 0,horizontally then the graph by c, and of a vertically(x + c) + byd dis. obtained (Remember by shiftingshifting the the⇥ graph graph of of axax⇤ 2 horizontallyhorizontally by by cc,, and and vertically vertically by by dd.. (Remember (Remember thatthatd>d>0meansmovingup,0meansmovingup,d<d<0meansmovingdown,0meansmovingdown,c>c>0meansmoving0meansmoving thatleft,andd>c<0meansmovingup,0meansmovingd<right0meansmovingdown,.) c>0meansmoving leftleft,and,andc<c<0meansmoving0meansmovingrightright.).) 2 If a =0,thegraphofafunctionf(x)=a(x + c) 2+ d is called a parabola. If a =0,thegraphofafunctionf(x)=a(x + c)2 + d is called a parabola. 6 2 TheIf a point=0,thegraphofafunction⇤ ( c, d) 2 is called thefvertex(x)=aof(x the+ c parabola.) + d is called a parabola. The point⇤ ( c, d) R2 is called the vertex of the parabola.
    [Show full text]