MrE's Algebra Blog

Algebra: Chapter 13, Lesson 2, page 580.

More Solving Quadratic Equations

Solve a quadratic equation of the form `ax^2 = k`

Example: `-3x^2 + 7 = 0` becomes …

`-3x^2 = -7` or `x^2=7/3`

then `x = ±sqrt(7/3)`, don’t forget to rationalize this to `x = ±sqrt(21)/3`!

Solve a quadratic equation by factoring one expression into a binomial square [of the form `(x + a)^2 = k`]

Example: `(x-5)^2 = 9`, if you take the square root of both sides becomes …

`x-5 = ±sqrt(9)` or `x = 5 ±sqrt(9)` or `x = 5 ± 3` or `x=8` or `x=2`

See here for more examples.

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: CST Review – Area of a Parallelogram, Triangle and Trapezoid

Definitions: `b` is the base and `h` is the height of the geometric figure below. Remember too, that the base and the height are at RIGHT ANGLES to each other!

Area, of a Triangle, `A = (b*h)/2` which is the same thing as `(1/2) * b * h`

Area of a Parallelogram, `A= b*h`.

Area of a Trapezoid, `A = (1/2) * h * (b_1 + b_2)` which is the same thing as `[h * (b_1 + b_2)]/2`

Algebra: Chapter 13, Lesson 1, page 576.

Introduction to Quadratic Equations

An equation that can be written in the form of `ax^2 + bx + c = 0` is a quadratic equation. This is the “standard form”. The solutions of the quadratic equation are called the:

• roots
• zeroes
• solutions
• x-intercepts
• answers

We can factor the equation and use the zero product property to find its solutions. There can be `0`, `1` or at most `2` solutions. This is because we have an exponent with power 2 (the `ax^2` term).

1. Equations of the form `ax^2 + bx = 0`

can be factored by taking out an `x` leaving us with `x(ax + b)` and using the zero product property we have `x = 0` and `ax + b = 0` as solutions.

2. Equations of the form `ax^2 + bx + c = 0`

can factored using the BOX METHOD with solutions again arrived at by using the zero product property.

Remember too, the inflexion point (the vertex, the point where the slope of the parabola changes direction) is `-b/(2a)`. This can help with the graph if you need to make one.

Purplemath has this link that is pretty good with more explanations and review.

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: CST Review – Area of a Circle

Area, A `A = π* (r^2)` or `A=πr^2` where `r`  is the radius of a circle, remember that and you can’t go wrong!

`π ≃ 3.14` or `π ≃ 22/7`

Remember, the relationship between the radius and the diameter,

`2r = d` or `r = d/2`

and that the circumference of a circle, `C` is

`C = 2 *π *r`  or  `C = 2πr`

Algebra: Chapter 11 BENCHMARK!

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Algebra 1a: CST Review – X-Y Graphing and more square root review!

Remember to graph, x (sideways) and y (vertically) and you’ll be OK!

Algebra: Chapter 12, Lesson 4, page 552.

Quadratic Functions

A quadratic function is defined by `f(x) = ax^2 + bx + c`. To make it easier, just replace the `f(x)` with `y` and treat as you have done in the past.

With a function in standard form `ax^2 + bx + c`, the vertex is `-b/(2a)` and the axis of symmetry is `x = -b/(2a)`. The vertex is the point on the PARABOLA where the slope changes sign from positive to negative or vice-versa. The axis of symmetry, you recall, is the line when the parabola can be “flipped” or “folded over” and still be symmetrical in shape.

You should plot at least 5 points when making a graph of the equation and DO NOT use a ruler to connect the points, this is a parabola, NOT a linear equation.

Where the parabola crosses the x-axis are called the ROOTS of the equation. These are also the x-intercepts!

ROOTS, ZEROES, X-INTERCEPTS and ANSWERS ALL MEAN THE SAME THING!

You can find the ROOTS easily by setting y = 0 and solving the quadratic equation with the factoring techniques from Chapter 6 and 10

OR

making a graph and seeing the 2 points (up to actually) where the parabola crosses the x-axis.

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: CST Review

Square Roots: The `sqrt(25) = 5`.

The `sqrt(5)` can be approximated as somewhere between the `sqrt(4)` and the `sqrt(9)`.

`sqrt(9)=3`

`sqrt(5) = ?`

`sqrt(4)=2`, so `sqrt(5)` is closer to `sqrt(4)` than `sqrt(9)`.  A good guess could be 2.2?

Pythagorean Theorem: Only works for right – triangles. The 3 sides of the triangle are `a`, `b`, and `c`. `a` and `b` are the short sides and `c`, the side OPPOSITE the right angle is called the hypotenuse.

The theorem states:

`c^2= a^2 + b^2` (if you know the 2 smaller sides, `a` and `b`)

or

`a^2 = c^2 – b^2` (if you know the hypotenuse, `c` and side `b`)

or

`b^2 = c^2 – a^2` (if you know the hypotenuse, `c` and side `a`)

Algebra: Chapter 12, Lesson 1 and Lesson 2, page 536 and 542.

Relations and Functions and Graphs

A relation is a set or ordered pairs. The domain of a relation is the set of first coordinates (the x’s). The range is the set of second (or y) coordinates.

A function is a relation that assigns one member of the domain EXACTLY one member of the range. In other words, one x can have ONLY one y value.

You can evaluate a function `f(x)=2x^2 + 5` for `f(2)` by substituting 2 for every x that you see. In our example then, `f(2)=2⋅4 + 5 = 13`

You can recognize a function with the vertical line test. You can find the domain and range of a function by seeing where the function is not defined. Good examples are `−`, we can’t take square roots of negative numbers and we cannot have `1/0` because that results in infinity or an undefined solution.

Here again, is a great link for Purplemath.

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Algebra 1a: CST Review

Square Roots: The `sqrt(25) = 5`.

The `sqrt(5)` can be approximated as somewhere between the `sqrt(4)` and the `sqrt(9)`.

`sqrt(9)=3`

`sqrt(5) = ?`

`sqrt(4)=2`, so `sqrt(5)` is closer to `sqrt(4)` than `sqrt(9)`.  A good guess could be 2.2?

Pythagorean Theorem: Only works for right – triangles. The 3 sides of the triangle are `a`, `b`, and `c`. `a` and `b` are the short sides and `c`, the side OPPOSITE the right angle is called the hypotenuse.

The theorem states:

`c^2= a^2 + b^2` (if you know the 2 smaller sides, `a` and `b`)

or

`a^2 = c^2 – b^2` (if you know the hypotenuse, `c` and side `b`)

or

`b^2 = c^2 – a^2` (if you know the hypotenuse, `c` and side `a`)

Algebra: Chapter 11 Review!

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Algebra 1a: CST Review

Square Roots: The `sqrt(25) = 5`.

The `sqrt(5)` can be approximated as somewhere between the `sqrt(4)` and the `sqrt(9)`.

`sqrt(9)=3`

`sqrt(5) = ?`

`sqrt(4)=2`, so `sqrt(5)` is closer to `sqrt(4)` than `sqrt(9)`.  A good guess could be 2.2?

Pythagorean Theorem: Only works for right – triangles. The 3 sides of the triangle are `a`, `b`, and `c`. `a` and `b` are the short sides and `c`, the side OPPOSITE the right angle is called the hypotenuse.

The theorem states:

`c^2= a^2 + b^2` (if you know the 2 smaller sides, `a` and `b`)

or

`a^2 = c^2 – b^2` (if you know the hypotenuse, `c` and side `b`)

or

`b^2 = c^2 – a^2` (if you know the hypotenuse, `c` and side `a`)

SNOW DAY #5

Algebra: Chapter 11, Lesson 9, page 519.

Equations with Radicals

A radical equation contains a variable in the radicand. To solve radical equations, first convert them to equations without radicals.

Just like taking the square root of 2 sides of an equation, you can also SQUARE both sides of an equation. This is helpful when you have radicals (square roots), because the square of a square root is just the thing in the radical!

For example:

`sqrt(2x) – 4 = 7`

first add 4 to both sides of the equation

`sqrt(2x) = 11`

now square both sides

`(sqrt(2x))^2 = 11^2`

and finally

`2x = 121` dividing by 2 yields the final answer `x = 121/2`

Don’t forget too, that sometimes you can have extraneous solutions, i.e., the answer doesn’t work so … you SHOULD check your answers by plugging them in and see if they WORK!

Click here for more information and examples!

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: Chapter 9, Lesson 4, page 413.

Inequalities and Absolute Value

If the inequality with absolute values looks like: `| A | < b`, then we solve the conjunction `-b < A < b`. Think of a number line, and the solution will be within the bounds of `-b` and `b`. This also works with `≤`.

If the inequality with absolute values look like: `| A | > b`, then we solve the disjunction `A < -b` OR `A > b`. On the number line, these solutions look like arrows on the outside of the values `-b` and `b`. This works for `≥` as well.

Click here for some examples!

Two of tonight’s homework problems solved by MrE are here! Just click it

Algebra: Chapter 11, Lesson 7 and Lesson 8, page 509 and 514.

Theorem of Pythagoras and its Uses

`c^2 = a^2 + b^2`,

where `a`, `b` and `c` are the sides of a RIGHT TRIANGLE. `a` and `b` are considered the sides of the triangle where `c` (opposite of the right angle) is called the hypotenuse.

The distance formula is a derivation of the Pythagorean Theorem and you can use it to find the distance from one coordinate point to another. Just remember, like in the slope equation, DON’T MIX UP THE DIRECTION that you define as `x_1` and `y_1`, they both go the same way!

`d = sqrt((x_1 – x_2)^2 + (y_1 – y_2)^2)`

Here is another quick explanation of the Theorem of Pythagoras and the Distance Formula.

Use a calculator with a `sqrt` key, it will make the word problems much easier to do!

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra: Chapter 9, Lesson 4, page 413.

Inequalities and Absolute Value

If the inequality with absolute values looks like: `| A | < b`, then we solve the conjunction `-b < A < b`. Think of a number line, and the solution will be within the bounds of `-b` and `b`. This also works with `≤`.

If the inequality with absolute values look like: `| A | > b`, then we solve the disjunction `A < -b` OR `A > b`. On the number line, these solutions look like arrows on the outside of the values `-b` and `b`. This works for `≥` as well.

Click here for some examples!

Two of tonight’s homework problems solved by MrE are here! Just click it

Algebra: Chapter 11, Lesson 7 and Lesson 8, page 509 and 514.

Theorem of Pythagoras and its Uses

`c^2 = a^2 + b^2`,

where `a`, `b` and `c` are the sides of a RIGHT TRIANGLE. `a` and `b` are considered the sides of the triangle where `c` (opposite of the right angle) is called the hypotenuse.

The distance formula is a derivation of the Pythagorean Theorem and you can use it to find the distance from one coordinate point to another. Just remember, like in the slope equation, DON’T MIX UP THE DIRECTION that you define as `x_1` and `y_1`, they both go the same way!

`d = sqrt((x_1 – x_2)^2 + (y_1 – y_2)^2)`

Here is another quick explanation of the Theorem of Pythagoras and the Distance Formula.

Use a calculator with a `sqrt` key, it will make the word problems much easier to do!

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: Chapter 9, Lesson 3, page 411.

Equations and Absolute Value

To solve an equation of the form `| A | = b`, solve the disjunction `A = b` OR `A = −b`. You will have 2 equations to solve with the right side of the second equation having the opposite sign of the first equation’s right side.

REMEMBER by definition, the solution of `| A | ≠ a` NEGATIVE NUMBER! So … the solution to these type of problems is the NULL SET! or the symbol `∅` !

Here is a link to examples!

Two of tonight’s homework problems solved by MrE are here! Just click it!