Quadratics: 3x^2 + 11x - 20 = 0, x^2 + 3x - 4 = 0, etc

[rachy7]

Can someone help me make sure this is correct?

A. Solve the following quadratic equations. Make sure to show all your work. Do not use any method (e.g., factoring, completing the square, quadratic formula, graphing) more than twice. Use the graphing method at least once.
1. 3x2 + 11x – 20 = 0
2. x2 + 3x – 4 = 0
3. 3x2 – x – 1 = 0

1. 3x^2 + 11x - 20 = 0

x = [-b + (b^2 - 4ac)^.5] / 2a

x = [-11 + (11^2 - 4*3*-20)^.5] / (2*3)

x = [-11 + (121 +240)^.5] / 6

x = [-11 + (361)^.5] / 6

x = (-11 + 19 ) / 6

x = 5, 8/6


2. x^2 + 3x - 4 = 0

x = -5, y = 6
x = -4, y = 0
x = -3, y = -4
x = -2, y = -6
x = -1, y = -6
x = 0, y = -4
x = 1, y = 0
x = 2, y = 6

The graph will show the coordinates to see that x = -4, 1




3. 3x^2 - x - 1 = 0

3x^2 - x - 1 = 0

3x^2 - x = 1

x^2 - x/3 = 1/3

x^2 - x/3 + 1/36 = 1/3 + 1/36

(x - 1/6)^2 = 13/36

x - 1/6 = + (13 /36)^.5

x = 1/6 + (13^.5) / 6

x = (1 - 13^.5) / 6, (1 + 13^.5) / 6,



B. State the different methods you used to solve each equation. Make sure that your work demonstrates both algebraic and graphing methods.

1. Quadratic Equation
2. Figure out wheat the graph would look like
3. Completing the Square

 

[galactus]

Re: Quadratics

Everything looks great, except for a minor sign problem.

1. 3x^2 + 11x - 20 = 0

x = [-b + (b^2 - 4ac)^.5] / 2a

x = [-11 + (11^2 - 4*3*-20)^.5] / (2*3)

x = [-11 + (121 +240)^.5] / 6

x = [-11 + (361)^.5] / 6

x = (-11 + 19 ) / 6

x = -5, 4/3