1. How many solutions exist for a quadratic equation? How do we determine algebraically whether the solutions are real or complex?
There can be 0, 1 or 2 solutions to any quadratic equation. Look at the discriminant: if b^2 - 4ac ? 0, then the roots are real; if b^2 - 4ac < 0, then the roots are complex.
2. What three techniques can be used to solve a quadratic equation? Demonstrate these techniques on the equation "x2 - 10x - 39 = 0".
1: x^2 - 10x - 39 = 0 can be solved by factoring: (x - 13)(x + 3) = 0
2: It can be solved by using the Quadratic Formula: x = [-(-10) ± ?(10^2 - 4(1)(-39)]/(2 • 1) etc.
3: It can be solved by completing the square: x^2 - 10x = 39
x^2 - 10x + 25 = 39 + 25
(x - 5)^2 = 64
x - 5 = ± 8 etc.
3. Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is four times its width; if the area of the garden is 196 square meters, what are its dimensions?
Let x = width and 4x = length
A = LW = (4x)(x) = 196
4x^2 = 196
x^2 = 49 etc.
4. Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer.
a) x^2 + 3x - 15 = 0: b^2 - 4ac = 3^2 - 4(1)(-15) = 9 - (-60) = 69 2 irrational solutions
b) x^2 + x + 4 = 0: 1^2 - 4(1)(4) = 1 - 16 = -15 2 complex solutions
c) x^2 – 4x - 7 = 0: 4^2 - 4(1)(-7) = 16 - (-28) = 44 same as (a)
d) x^2 – 8x + 16 = 0: (-8)^2 - 4(1)(16) = 64 -64 = 0 1 double rational root
e) 2x^2 - 3x + 7 = 0: (-3)^2 - 4(1)(7) = 9 - 28 = -19 same as (a)
f) x^2 – 4x - 77 = 0: (-4)^2 - 4(1)(-77) = 16 - (-308) = 324 2 rational solutions
g) 3x^2 - 7x + 6 = 0: (-7)^2 - 4(3)(6) = 49 - 72 = -23 same as (b)
h) 4x^2 + 16x + 16 = 0: 16^2 - 4(4)(16) = 256 - 256 = 0 same as (d)
5. Find an equation for which -3 and 4 are solutions.
(x - (-3)) (x - 4) = 0
(x + 3)(x - 4) = 0
x^2 - x - 12 = 0
6. What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.
The solutions will be complex conjugates. x^2 + 1 = 0 0^2 - 4(1)(1) = 0 - 4 = -4
x^2 = -1 x = ±i
7. Create a real-life situation that fits into the equation (x + 3)(x - 5) = 0 and express the situation as the same equation.
That's one that is difficult for which to make up a problem.
