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Algebra Basics: What Is Algebra? - Math Antics
By Lumos Learning
This video gives an overview of Algebra and introduces the concepts of unknown values and variables. It also explains that multiplication is implicit in Algebra.
Work Word Problems | MathHelp.com
By MathHelp.com
To solve a polynomial inequality, like the one shown here, our first step is to write the corresponding equation. In other words, we simply change the inequality sign to an equals sign, and we have x^2 – 3 = 9 – x. Next, we solve the equation. Since we have a squared term, we first set the equation equal to 0. So we move the 9 – x to the left side by subtracting 9 and adding x to both sides of the equation. This gives us x^2 + x – 12 = 0. Next, we factor the left side as the product of two binomials. Since the factors of negative 12 that add to positive 1 are positive 4 and negative 3, we have x + 4 times x – 3 = 0. So either x + 4 = 0 or x – 3 = 0, and solving each equation from here, we have x = -4, and x = 3. Now, it’s important to understand that the solutions to the equation, -4 and 3, represent what are called the “critical values” of the inequality, and we plot these critical values on a number line. However, notice that our original inequality uses a greater than sign, rather than greater than or equal to sign, so we use open dots on our critical values of -4 and positive 3. Remember that ‘greater than’ or ‘less than’ means open dot, and ‘greater than or equal to’ or ‘less than or equal to’ means closed dot. Now, we can see that our critical values have divided the number line into three separate intervals: less than -4, between -4 and 3, and greater than 3. And here’s the important part. Our next step is to test a value from each of the intervals by plugging the value back into the original inequality to see if it gives us a true statement. So let’s first test a value from the “less than -4” interval, such as -5. If we plug a -5 back in for both x’s in the original inequality, we have -5 squared – 3 greater than 9 minus a -5, which simplifies to 25 – 3 greater than 9 + 5, or 22 greater than 14. Since 22 greater than 14 is a true statement, this means that all values in the interval we’re testing are solutions to inequality, so we shade the interval. Next, we test a value from the “between -4 and 3” interval, such as 0. If we plug a 0 back in for both x’s in the original inequality, we have 0 squared – 3 greater than 9 – 0, which simplifies to 0 – 3 greater than 9, or -3 greater than 9. Since -3 greater than 9 is a false statement, this means that all values in the interval we’re testing are not solutions to inequality, so we don’t shade the interval. Next, we test a value from the “greater than 3” interval, such as 4. If we plug a 4 back in for both x’s in the original inequality, we have 4 squared – 3 greater than 9 – 4, which simplifies to 16 – 3 greater than 5, or 13 greater than 5. Since 13 greater than 5 is a true statement, this means that all values in the interval we’re testing are solutions to inequality, so we shade the interval. Finally, we write the answer that’s shown on our graph in set notation. The set of all x’s such that x is less than -4 or x is greater than 3.
12 - Solving 3-Variable Linear Systems of Equations - Substitution Method
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
01 - Visualize Roots of Equations - Linear, Quadratic, Cubic, Quartic Solutions
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
Buffer Solutions
By The Organic Chemistry Tutor
This chemistry video tutorial explains how to calculate the pH of a buffer solution using the henderson hasselbalch equation. It explains the concept, components, and function of a buffer solution. A buffer solution consist of a weak acid and its conjugate weak base counterpart. It's purpose is to maintain a relatively constant pH value. This video discusses the relationship between the pH and pKa values with the relative amounts of weak acid and weak base components in the buffer solution. This video contains plenty of examples and practice problems.
pH of Weak Acids and Bases - Percent Ionization - Ka & Kb
By The Organic Chemistry Tutor
This chemistry video explains how to calculate the pH of a weak acid and a weak base. It explains how to calculate the percent ionization of a weak acid using the acid dissociation constant Ka.
01 - Solving Equations in Quadratic Form - Part 1 (Learn to Solve Equations in Algebra)
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
10 - What are Imaginary Numbers?
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 1)
By Lumos Learning
Here is a great exam review video reviewing all of the main concepts you would have learned in the MPM1D grade 9 academic math course. The video is divided in to 3 parts. This is part 1: Algebra. The main topics in this section are exponent laws, polynomials, distributive property, and solving first degree equations. Please watch part 2 and 3 for a review of linear relations and geometry. If you watch all 3 parts, you will have reviewed all of grade 9 math in 60 minutes. Enjoy! Visit jensenmath.ca for more videos and course materials.
20 - The Rational Root Theorem, Part 1 (Rational Roots of Polynomials)
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
10th Grade Modeling with Systems of Equations
By
10th grade math lesson on modeling with systems of equations. The lesson involved high school modeling standard A.CED.A.2 from the MA Curriculum Frameworks for Mathematics, which specifies creating equations in two or more variables.