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Ch8 Quadratic Equation Solving Methods

Ch8 Quadratic Equation Solving Methods. General Form of Quadratic Equation a x 2 + b x + c = 0 A quadratic Equation: x 2 – 7x + 10 = 0 a = _____ b = _____ c = ______ Methods & Tools for Solving Quadratic Equations Factor

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Ch8 Quadratic Equation Solving Methods

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  1. Ch8 Quadratic Equation Solving Methods • General Form of Quadratic Equation • ax2 + bx + c = 0 • A quadratic Equation: x2 – 7x + 10 = 0 a = _____ b = _____ c = ______ • Methods & Tools for Solving Quadratic Equations • Factor • Apply zero product principle (If AB = 0 then A = 0 or B = 0) • Square root method • Completing the Square • Quadratic Formula 1 -7 10 Example1:Example 2: x2 – 7x + 10 = 0 4x2 – 2x = 0 (x – 5) (x – 2) = 0 2x (2x –1) = 0 x – 5 = 0 or x – 2 = 0 2x=0 or 2x-1=0 + 5 + 5 + 2 + 2 2 2 +1 +1 2x=1 x = 5 or x = 2 x = 0 or x=1/2

  2. 8.1 Square Root Method • Solving a Quadratic Equation with the Square Root Method • Example 1: Example 2: • 4x2 = 20 (x – 2)2 = 6 • 4 • x – 2 = +6 • x2 = 5 + 2 + 2 • x = + 5 x = 2 + 6 • So, x = 5 or - 5 So, x = 2 + 6 or 2 - 6 You try one. What is different about this one? : 2x2 + 1 = 0

  3. Completing the Square If x2 + bx is a binomial then by adding b2 which is the square of half 2 the coefficient of x, a perfect square trinomial results: x2 + bx + b2 = x + b2 2 2 Solving a quadratic equation with ‘completing the square’ method. Example: Step1: Isolate the Binomial x2 - 6x + 2 = 0 -2 -2 Step 2: Find ½ the coefficient of x (-3 ) x2 - 6x = -2 and square it (9) & add to both sides. x2 - 6x + 9 = -2 + 9 (x – 3)2 = 7 x – 3 = + 7 x = (3 + 7 ) or (3 - 7 ) Note: If the coefficient of x2 is not 1 you must divide by the coefficient of x2 before completing the square. ex: 3x2 – 2x –4 = 0 (Must divide by 3 before taking ½ coefficient of x) Step 3: Apply square root method

  4. 8.2 Quadratic Formula General Form of Quadratic Equation: ax2 + bx + c = 0 Quadratic Formula:x = -b + b2 – 4ac discriminant: b2 – 4ac 2a if 0, one real solution if >0, two unequal real solutions if <0, imaginary solutions Solving a quadratic equation with the ‘Quadratic Formula’ 2x2–6x + 1= 0 a = ______ b = ______ c = _______ x = - (-6) + (-6)2 – 4(2)(1) 2(2) = 6 + 36 –8 4 = 6 + 28 = 6 + 27 = 2 (3 + 7 ) = (3 + 7 ) 4 4 4 2 2 -6 1

  5. 8.5 & 8.6 Quadratic Functions & Graphs y = x2 • x y • 0 0 • 1 • -1 1 • 4 • -2 4 Vertex Lowest point if The parabola opens upward, And highest point if parabola Opens downward. A Parabola Do you know what an axis of symmetry is?

  6. Quadratic Functions & Graphs y = x2 - 2 • x y • 0 -2 • -1 • -1 -1 • 2 • -2 2 Vertex Notice this graph is shifted down 2 from the origin. Y = x2 – k (shifts the graph down k units) Y = x2 + k (shifts the graph up k units) To shift the graph to the right or to the left y = (x – h)2 (shifts the graph to the right) y = (x + h)2 (shifts the graph to the left)

  7. General Form of a Quadratic y = ax2 + bx + c a, b, c are real numbers & a 0 A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______ 1 4 3 x y -5 8 -4 3 -3 0 -2 -1 -1 0 0 3 1 8 Formula for Vertex: X = -b 2a Plug x in to Find y Where is the vertex? Where is the axis of symmetry?

  8. The Role of “a”

  9. Quadratic Equation Forms • Standard Form: • Vertex Form: Vertex = (h, k)

  10. Examples Find the vertex, axis of symmetry, and graph each a. b. c. Vertex (5, 2) Vertex (-2, -3) Vertex (4, -6)

  11. Convert from Vertex Form to Standard Form Vertex Form: y = 2(x + 2)2 + 1 To change to standard form, perform multiplication, add, and combine like terms. y = 2 (x + 2) (x + 2) + 1 y = 2 (x2 + 2x + 2x + 4) + 1 y = 2 (x2 + 4x + 4) + 1 y = 2x2 + 8x + 8 + 1 y = 2x2 + 8x + 9 (Standard Form)

  12. Convert from Standard Form to Vertex Form(Completing the Square – Example 1) Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, factor out the coefficient from x2 and x terms. Step 2: Calculate the value of : (b/2)2 Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2 Step 4: Factor & Simplify Example 1: y = x2 –6x – 1 (Standard Form) (b/2)2 = (-6/2)2 = (-3)2 = 9 y = (x2 –6x + 9) – 1 -9 y = (x – 3) (x – 3) – 1 – 9 y = (x – 3)2 – 10 (Vertex Form)

  13. Convert from Standard Form to Vertex Form(Completing the Square – Example 2) Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, factor out the coefficient from the x2 and x term. Step 2: Calculate the value of : (b/2)2 Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2 Step 4: Factor & Simplify Example 2: y = 2x2 +4x – 1 (Standard Form) y = 2( x2 + 2x) –1 (2/2)2 = (1)2 = 1 y = 2(x2 +2x + 1) – 1 -2 (WHY did we subtract 2 instead of 1?) y = 2(x + 1) (x + 1) – 1 – 2 y = 2(x + 1)2 – 3 (Vertex Form)

  14. Solving Quadratic Equations General Form of a Quadratic Equation y = ax2 + bx + c 0 = ax2 + bx + c (If y = 0, we can solve for the x-intercepts) A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______ 1 4 3 Graphical Solution Numerical Solution Symbolic/Algebraic Solution x2 + 4x + 3 = 0 (x + 3) (x + 1) = 0 x + 3 = 0 x + 1 = 0 x = -3 x = -1 x y -5 8 -4 3 -3 0 -2 -1 -1 0 0 3 1 8 Vertex (-2, -1) Formula for Vertex: X = -b 2a

  15. Number of Solutions y = x2 + 4x + 3 y = x2 - 4x + 4 y = 2x2 + 1 1 Real Solution x = 2 NO Real Solutions (No x-intercepts) 2 Real Solutions x = -3 x = -1

  16. An Application 3 in Height of a picture = X Length of picture = X + 5 The frame is 3 inches thick on all sides. If the overall area of the picture and frame is 336 sq inches, find the dimensions (height and length) of the picture. x 3 3 x + 5 3 in. Arearectangle = Length x Height Length = x + 5 + 6 = x + 11 Height = x + 6 336 = (x + 6) (x+11) 336 = x2 +11x +6x +66, so, x2 +17x – 270 = 0 (x – 10) (x + 27) = 0 x = 10 or x = -27 So, the picture is 10 inches by 15 inches

  17. Another Application Two cars left an intersection at the same time, one heading Due north, the other due west. Some time later they were exactly 100 Miles apart. The car headed north had gone 20 miles farther than the car Headed west. How far had each car traveled? (P. 269) Leg12 + leg22 = Hypotenuse2 North X2 + (x+20)2 = 1002 X2 + X2 + 40x + 400 = 10000 2X2 + 40x + 400 = 10000 2X2 + 40x - 9600 = 0 X2 + 20x - 4800 = 0 (x + 80) (x – 60) = 0 X = -80 or x = 60 100 X + 20 West x Intersection

  18. Things to Know for the Quiz • Standard Form of a quadratic: y = ax2 + bx + c • Vertex Form of a quadratic: y = a(x – h)2 + k • Find and graph the vertex • Use an x/y chart to plot points & graph the parabola • Show and give the equation for the line of symmetry. • Convert from standard to vertex form and vice vera • 2. Solving a quadratic equation to find its x-intercepts using these methods: • Factor and apply zero product principle • Square Root Method • Completing the Square • Quadratic Formula • 3. Formulas I will give you: • Quadratic Formula • Formula for vertex of a standard form quadratic: x = -b • 2a Use when the equation won’t factor easily

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