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C2 st lecture 2 handout

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C2 st lecture 2 handout

  1. 1. Lecture 2 - Quadratic Equations and Straight Lines C2 Foundation Mathematics (Standard Track) Dr Linda Stringer Dr Simon Craik l.stringer@uea.ac.uk s.craik@uea.ac.uk INTO City/UEA London
  2. 2. Lecture 2 skills factorize a quadratic equation complete the square use the quadratic formula simplify a surd (using a calculator if necessary) sketch a straight line, given the equation find the equation of a line, given the gradient and a point find the equation of a line, given two points find the midpoint of two points find the distance between two points
  3. 3. Lecture 2 vocabulary and symbols quadratic surd √ square root gradient intercept > ≥ < ≤ ∞
  4. 4. Quadratic equations A quadratic equation is an equation that involves a variable squared, for example x2 + 4x + 3 = 0 3x2 + 12x − 1 = 0 We have three methods for solving quadratic equations Factorising Completing the square Using the quadratic formula You need to be able to use ALL THREE METHODS
  5. 5. Solving equations equal to zero If a × b = 0 then a = 0 or b = 0. For example If 5x = 0 then x = 0. If 5(x − 3) = 0 then (x − 3) = 0, so x = 3. If (x − 3)(x + 4) = 0 then (x − 3) = 0 or (x + 4) = 0. It follows that x = 3 or x = −4.
  6. 6. Factorising when the coefficient of x2 is 1 Change the quadratic equation x2 + bx + c = 0 into the form (x + b1)(x + b2) = 0 where b1 and b2 are numbers such that b1 + b2 = b and b1 × b2 = c Since (x + b1)(x + b2) = x2 + b2x + b1x + b1b2 = x2 + x(b2 + b2) + b1b2 = x2 + bx + c If (x + b1)(x + b2) = 0, then (x + b1) = 0 or (x + b2) = 0. It follows that x = −b1 or x = −b2
  7. 7. Factorising a quadratic equation - example 1 Question: Solve by factorising x2 + 8x + 7 = 0 Answer: Find b1 and b2 such that b1 + b2 = 8 and b1 × b2 = 7 Clearly b1, b2 = 7, 1 (x + 7)(x + 1) = 0 Either (x + 7) = 0 or (x + 1) = 0, so x = −7 and x = −1. Substitute our solutions into our original equation to check. (−7)2 + (8 × (−7)) + 7 = 0. (−1)2 + (8 × (−1)) + 7 = 0
  8. 8. Factorising when the coefficient of x2 is not 1 To solve ax2 + bx + c = 0 find two numbers b1 and b2 such that b1 + b2 = b and b1 × b2 = a × c then split the middle term ax2 + bx + c = ax2 + b1x + b2x + c and finally factorise the first two terms and the last two terms (see example 2)
  9. 9. Factorising a quadratic equation - example 2 Question: Solve by factorising 6x2 + 19x + 10 = 0 Answer: Find b1, b2 such that b1 + b2 = 19 and b1 × b2 = 6 × 10 = 60 Clearly b1, b2 = 15, 4 First split the middle term 6x2 + 19x + 10 = 6x2 + 15x + 4x + 10 Now factorise the first two terms and the last two terms = 3x(2x + 5) + 2(2x + 5) = (3x + 2)(2x + 5) So (3x + 2)(2x + 5) = 0 It follows that (3x + 2) = 0 or (2x + 5) = 0. If (3x + 2) = 0, then x = −2 3 If (2x + 5) = 0, then x = −5 2 . The solutions are x = −2 and x = −5 .
  10. 10. Factorising when the coefficient of x2 is not 1 ALTERNATIVE METHOD Change a quadratic equation of the form ax2 +bx+c = 0 into the form (a1x+c1)(a2x+c2) = 0 Find all the pairs of factors of a (call them a1, a2), and all the pairs of factor of c (call them c1, c2). Experiment to see which combination a1c2 + a2c1 = b. Solve 15x2 − 14x − 8 = 0
  11. 11. Completing the square Al-Khwarizmi - the father of algebra
  12. 12. Completing the square, x2 + bx + c = 0 Check the coefficient of x2 is 1: x2 + 6x + 2 = 0 Move c to the right side: x2 + 6x = −2 Add (b 2 )2 to both sides: x2 + 6x + (6 2)2 = −2 + (6 2 )2 Tidy up both sides: x2 + 6x + 9 = 7 Write the left side as (x + b 2 )2: (x + 3)2 = 7 Take the square root of both sides: x + 3 = ± √ 7 Move the constant to the right side: x = −3 ± √ 7
  13. 13. Completing the square - example 2 Question: Solve 4x2 − 2x − 3 = 0 by completing the square Answer: First divide by 4. x2 − 1 2 x − 3 4 = 0 Check the coefficient of x2 is 1: x2 − 1 2 x − 3 4 = 0 Move c to the right side: x2 − 1 2 x = 3 4 Add (b 2 )2 to both sides: x2 − 1 2 x + (−1 4 )2 = 3 4 + (−1 4)2 Tidy up both sides: x2 − 1 2 x + 1 16 = 13 16 Write the left side as (x + b 2 )2: (x − 1 4 )2 = 13 16 Take the square root of both sides: x − 1 4 = ± 13 16 Move the constant to the right side: x = 1 4 ± 13 16 The solution is x = 1± √ 13 4 .
  14. 14. Quadratic formula To solve ax2 + bx + c = 0, you will be given the quadratic formula x = −b ± √ b2 − 4ac 2a Question: Solve 5x2 + 9x − 3 = 0 using the quadratic formula Solution: x = −9 ± 92 − 4 × 5 × (−3) 2 × 5 So x = −9± √ 141 10 (surd form) = 0.29, −2.09 (to 2 d.p.)
  15. 15. Complete the square to prove the quadratic formula Take a general quadratic equation ax2 + bx + c = 0. First we divide through by a, so x2 + b a x + c a = 0. Move the constant term to the right x2 + b a x = −c a . Add ( b 2a )2 to both sides x2 + b a x + ( b 2a )2 = −c a + ( b 2a )2. Write the left side as a square (x + b 2a )2 = −c a + ( b 2a )2. Rearrange the right side to get (x + b 2a )2 = b2−4ac 4a2 . Take the square root of both sides x + b 2a = ± √ b2−4ac 2a Then x = −b± √ b2−4ac 2a as required.
  16. 16. Quadratic formula - example 2 Solve 3x2 − 8x + 2 = 0. Answer: a = 3, b = −8, c = 2. By the formula we have x = −(−8) ± (−8)2 − 4 × 3 × 2 2 × 3 Simplifying x = 8 ± √ 64 − 24 6 It follows x = 8 ± √ 40 6 and so our solutions are x = 4 + √ 10 3 , x = 4 + √ 10 3
  17. 17. Quadratic formula - example 3 Solve x2 = 58x − 2. Answer: First rearrange the equation x2 − 58x + 2 = 0 a = 1, b = −58, c = 2. By the formula we have x = −(−58) ± (−58)2 − 4 × 1 × 2 2 × 1 Simplifying x = 58 ± √ 3364 − 8 2 It follows x = 58 ± √ 3356 2 and so our solutions are x = 58 − √ 839, x = 58 + √ 839
  18. 18. Surds Surds must be presented in a simplified form. Your calculator will simplify surds for you! There must be no square factor under the square root sign. To simplify use the rule √ a × b = √ a × √ b Simplify √ 50√ 50 = √ 2 × 25 = √ 2 × √ 25 = 5 √ 2. √ 10 can not be simplified.
  19. 19. Cartesian coordinates, (x, y) −6 −4 −2 2 4 6 −6 −4 −2 2 4 6 • (4, 2) • • • x y
  20. 20. The gradient of a straight line Gradient (slope) = Vertical change Horizontal change = Change in y Change in x = ∆y ∆x = Rise Run ∆x (run) ∆y (rise) x y For each step to the right, the gradient tells you how many steps up (or down)
  21. 21. The gradient of a straight line −4 −2 2 4 −4 −2 2 4 x y −4 −2 2 4 −4 −2 2 4 x y −4 −2 2 4 −4 −2 2 4 x y −4 −2 2 4 −4 −2 2 4 x y
  22. 22. A straight line - the equation, gradient and y-intercept A straight line has equation y = mx + c c is the y-intercept. This is the y-coordinate of the point where the line passes through the y-axis. The line intercepts the y-axis at the point (0, c). m is the gradient of the line - for every one step to the right, you go m steps upwards A steep line has a large gradient (m > 1 or m < −1.) A shallow line has a small gradient (−1 < m < −1.) A horizontal line has gradient 0. In this case the equation of the line is y = c. A vertical line has gradient ∞. In this case the equation of the line is x = a (for some constant a.)
  23. 23. Sketch a line with equation y = x + 2 −6 −4 −2 2 4 6 −6 −4 −2 2 4 6 • • x y Find the y-intercept Let x = 0 Then y = 0 + 2 = 2 The line crosses the y-axis at (0, 2) Find the x-intercept Let y = 0 Then 0 = x + 2, so x = −2 The line crosses the x-axis at (−2, 0) Plot the intercepts Check the gradient Sketch the line
  24. 24. Sketch a line To sketch the line with equation y = mx + c Find the coordinates of two points on the line - usually it is best to find the y-intercept and x -intercept. Plot the two points. Check the gradient. Sketch the line
  25. 25. A line defined by a gradient and a point A straight line can also be defined by a gradient m and a point (x0, y0) through which the line passes. To get the equation of the line substitute x0 and y0 into the formula y = mx + c to find c. Question: Find the equation of the straight line with gradient 0.5 that passes through (6, 2). Answer: Substitute 2 = 0.5 × 6 + c. So 2 = 3 + c and c = −1. The equation is y = 0.5x − 1 or equivalently y = 1 2 x − 1
  26. 26. A line defined by two points A straight line can also be defined by 2 points (x0, y0) and (x1, y1) through which the line passes. To get the equation of the line we need to find the gradient m and the y-intercept c. The gradient is given by the formula: m = y1 − y0 x1 − x0 To find c we substitute in one of our points, say (x0, y0) into the line equation y = mx + c.
  27. 27. A line defined by two points - example Question: Find the equation of the straight line passing through the points (1, 1) and (−1, 3). Answer: First find the gradient m = 3 − 1 −1 − 1 = 2 −2 = −1 Substitute one of the points into the line equation to find c. We have 1 = −1 × 1 + c so c = 2. The equation of our line is y = −x + 2. We should now check our answer. Substitute in the other point (−1) × (−1) + 2 = 3.
  28. 28. Midpoint To find the mid-point of two points (x0, y0) and (x1, y1) we calculate the midpoint of x0 and x1 and the midpoint of y0 and y1. The midpoint is x0 + x1 2 , y0 + y1 2 Question: What is the midpoint of the points (1, 1) and (−1, 3)? Answer: The midpoint is 1 + (−1) 2 , 1 + 3 2 = (0, 2)
  29. 29. Distance To find the distance between two points (x0, y0) and (x1, y1) we use the formula (x1 − x0)2 + (y1 − y0)2 Question: What is the distance from (1, 1) to (−1, 3)? Answer: The distance is ((−1) − 1)2 + (3 − 1)2 = (−2)2 + 22 = √ 8 = 2 √ 2

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