College Algebra

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Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Find the zeros of quadratic functions.
- Solve quadratic equations, by using the principle of zero products, principle of square roots, by completing the square, and the quadratic formula.
- Solve an applied problem using a quadratic equation.


A quadratic equation in standard form looks like `ax^2+bx+c = 0`, where `a != 0` and a, b, and c are all real numbers.

And a quadratic function is of the form `f(x) = ax^2+bx+c`, where `a != 0` and a, b, and c are real numbers.


The roots of a quadratic equation are equivalent to the zeros of its associated quadratic function.

The Principle of Zero Products states that if `a*b = 0`, then `a = 0` or `b = 0`.

And the Principle of Square Roots states that if `x^2 = s` then `x = sqrt(s)` or `x = -sqrt(s)`.



Solve the quadratic equation.

`x^2-7x-30 = 0`. So we can solve this by factoring. `x*x = x^2`, and `+3*-10 = -30` and then if we check, we'll get `3x-10x = -7x`, which is our middle term.

By the Principle of Zero Products, `x+3 = 0` or `x-10 = 0`. So `x = -3` or `x = 10`.

Let's check `x = -3`. That means `(-3)^2-7(-3)-30` has to equal 0. So that's `9+21-30 = 0`, which gives us `0 = 0`, a true statement.

And then let's check `x = 10`. `(10)^2-7(10)-30`. Does that equal zero? We get `100-70-30`, which is `0 = 0`, another true statement.

So our roots are -3 and 10.



Solve by completing the square to obtain an exact solution.

`x^2-8x = 20`. In order to complete the square, we need to take the coefficient of our x term, divide by 2, square it, and add that value to both sides of our equation.

So `(-8/2)^2 = 16`. So we get `x^2-8x+16 = 20+16`. And then we can factor. `x*x = x^2` and `-4*-4` multiply to make 16, but add to make -8. So then we have 36 on the right. So this is the same as `(x-4)^2 = 36`, and if we square root both sides, we'll get `x-4 = +-6`.

So `x = 4+-6`. So `x = 10" or "-2`.

Now let's check `x = 10`. `10^2-8(10) = 20`. That means `100-80 = 20` or `20 = 20`, a true statement.

Now let's check `x = -2`. `(-2^2 -8(-2) = 20)` or `4+16 = 20` or `20 = 20`, a true statement.

So the roots of this quadratic equation are 10 and -2.



The method of completing the square can be used to solve any quadratic equation.

Let's use the method of completing the square on the standard quadratic equation to solve for x.

So if we subtract c from both sides, we get `ax^2+bx = -c`. And then if we divide all of our terms by a, we get `x^2+b/ax = -c/a`. And now if we complete the square, we have `x^2+b/ax+(b/(2a))^2 = -c/a+(b/(2a))^2. And let's go ahead and add our fractions on the right-hand side by making common denominators. If we multiply the top and bottom by 4a, we'll have `-(c*4a)/(4a)^2` plus and then square `b/(2a)` to get `b^2/(4a)^2`. Now the right-hand side becomes `(b^2-4ac)/(4a^2)`, and on the left-hand side, we can go ahead and factor it. So we have `(x+b/(2a))^2`. Now if we take the square root of both sides, we have `x+b/(2a) = +-sqrt(b^2-4ac)`, all over the `sqrt(4a^2)`, which is 2a. Now let's subtract `b/(2a)` from both sides. So we get `x = (-b+-sqrt(b^2-4ac))/(2a)`.

This is how we derive the quadratic formula. The quadratic formula is `x = (-b+-sqrt(b^2-4ac))/(2a)`.



Use the quadratic formula to solve the quadratic equation.

`5x^2+3 = 10x`. So we need to start by bringing our 10x over to the left side. So we get `5x^2-10x+3 = 0`. `a = 5`, `b = -10`, and `c = 3`.

So `x = (-b+-sqrt(b^2-4ac))/(2a)`. And let's just plug in `x = (-(-10)+-sqrt(100-4(5)(3)))/(2(5))`. So `x = (10+-sqrt(100-60))/10` or `x = (10+-sqrt(40))/10`. 40 can be broken down to `4*10`, and `4 = 2*2`, and 10 cannot be broken down any further.

So we have `(10+-2sqrt(10))/10`. And if we divide everything by 2, we get `(5+-sqrt(10))/5`.



For `ax^2+bx+c = 0`, where a, b, and c are all real numbers, if `b^2-4ac = 0`, our roots are real and equal. If `b^2-4ac > 0`, our roots are real and different. And if `b^2-4ac < 0`, our roots are imaginary.



The sum of the squares of two consecutive numbers is 481. Find the numbers.

So if we let `x = "first number"`, and `x+1 = "the next number"`, then the sum of the squares of the two consecutive numbers is `x^2+(x+1)^2` and it equals 481. So that means `x^2+x^2+2x+1 = 481` or `2x^2+2x+1 = 481`. And if we subtract 481 from both sides, we get `2x^2+2x-480 = 0`. Now let's divide everything by 2. That means `x^2+x-240 = 0`, and if we solve this using the quadratic formula, we'll use `a = 1`, `b = 1`, and `c = -240`.

So `x = (-b+-sqrt(b^2-4ac))/(2a)`. This equals `(-1+-31)/2`, which is `30/2` or `-32/2`, which is 15 or -16.

We can go ahead and cross -16 out here, because it doesn't apply. Consecutive numbers are understood to be integers which follow one other in the common scale of 1, 2, 3, 4, etc.. So the numbers are 15 and 16.