Solving Quadratics Graph: y = 2x 2 2 y = x 2 x 12 y = x 2 + 6x + 9 y = x 2 + 6x + 3 Roots are: real, rational real, rational real, rational, equal real, irrational 1
To find the roots algebraically, make sure each equation is set equal to 0 (all y values along the x axis are 0) and try to factor: 0 = 2x 2 2 0 = x 2 x 12 0 = x 2 + 6x + 9 0 = x 2 + 6x + 3 2
Quadratic Formula If the quadratic equation is unfactorable, you must use the quadratic formula to solve for the roots: Example 1: 0 = x 2 + 6x + 3 3
More examples with quadratic formula: 4
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Discriminant The discriminant will tell you the nature of the roots b 2 4ac y = 2x 2 2 If the discriminant is > 0 and a perfect square, then the roots are real and rational y = x 2 + 6x + 3 If the discriminant is > 0 and not a perfect square then the roots are real and irrational 7
y = x 2 + 6x + 9 If the discriminant is = 0, then the roots are real, rational and equal y = x 2 + x + 3 If the discriminant is < 0, then the roots are imaginary 8
y = x 2 x 12 a = 1, b = 1, c = 12 b 2 4ac ( 1) 2 4(1)( 12) 1 + 48 49 # is a perfect square, so roots are real and rational 9
b 2 4ac Positive Negative Zero There are two real roots. There are two x intercepts. The roots are unequal. There are two imaginary roots. There are no x intercepts. The roots are unequal. There is one real root. There is one x intercept. The roots are equal. * If it is positive, you must also include rational/irrational. If the number is a perfect square, then the roots are rational. If the number is not a perfect square, then the roots are irrational. 10
Describe the nature of the roots: b 2 4ac < 0 then the roots are imaginary = 0 then the roots are real, rational and equal > 0 and a perfect square, then the roots are real and rational > 0 and not a perfect square then the roots are real and irrational y = 2x 2 + 3x 4 y = x 2 + 4x + 4 y = x 2 + 5x 2 y = x 2 3x 2 y = x 2 4x 8 y = x 2 9 11
Complete the Square Another way to solve quadratic equations is to complete the square: Example 1: 3x 2 + 7 = 31 Easy with no b value! Example 2: 6x 2 4 = 38 12
Example 3: x 2 8x + 16 = 18 what is the connection between the b and c values? Example 4: x 2 2x + 1 = 10 Example 5: x 2 + 12x + 36 = 50 13
Vertex Form We are used to seeing parabolas written in standard form: y = ax 2 + bx + c It is easy to factor and find the roots this way. Sometimes we use vertex form, where we can easily find the vertex and the axis of symmetry. y = a(x h) 2 + k where (h, k) is the vertex and x = k is the axis of symmetry 14
1. Write equation as ax 2 + bx = c 2. Add (b/2a)2 to both sides to create a perfect square 3. Rewrite the perfect square in factored form ( ) 2 4. Move all terms to one side 15
1. Write equation as ax 2 + bx = c 2. Add (b/2a)2 to both sides to create a perfect square 3. Rewrite the perfect square in factored form ( ) 2 4. Move all terms to one side 16
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Sum and Product of Roots If y = x 2 + 5x + 6, find the sum and product of the roots. SUM b a PRODUCT c a 18
Without solving, find the sum and product of the roots of the equation: 2x 2 3x 2 = 0 19
Without solving, find the sum & product of the roots of the following equation: 9x 2 8x = 15 20
Write the quadratic equation given the following roots: 4 and 2 2 ways! 21
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Lines and Parabolas The most amount of times a line and a parabola can intersect is 2. They can also intersect just once, or no times. 0 1 1 2 23
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To solve an intersection algebraically use the substitution method: Set each equation equal to each other: Move all terms to one side of = and set equal to 0: y = x 2 3x 10 y = x + 2 Factor and solve for x: Plug each x value back into the linear equation and solve for y, leaving you with two points. y = x + 2 y = x + 2 25
To solve an intersection algebraically use the substitution method: Use the equation that is solved for a variable and plug into other eqn: Move all terms to one side of = and set equal to 0: y = x 2 6x + 1 y + 2x = 6 Factor and solve for x: Plug each x value back into the linear equation and solve for y, leaving you with two points. 26