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### C2 st lecture 2 handout

• 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. 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. Lecture 2 vocabulary and symbols quadratic surd √ square root gradient intercept > ≥ < ≤ ∞
• 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. 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. 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. 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. 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. 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. 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. Completing the square Al-Khwarizmi - the father of algebra
• 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. 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. 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. 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. 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. 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. 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. Cartesian coordinates, (x, y) −6 −4 −2 2 4 6 −6 −4 −2 2 4 6 • (4, 2) • • • x y
• 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. 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. 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. 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. 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. 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. 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. 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. 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. 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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