This week in Pre Calculus 11, we reviewed Rational Expressions in preparation for the quiz.

Example question:

X² + 3X – 10 ÷ X² – 5X + 6
3x²+13x-10    2x² + 6x – 36

We can turn the division symbol into multiplication by replacing the fraction on the right with its reciprocal.

X² + 3X – 10 .  2x² + 6x – 36
3x²+13x-10    X² – 5X + 6

Now, we factor:

(X + 5) (X – 2) .  (2x+12)(x-3)
(3x-2)(x+5)        (x-2)(x-3)

Now, we can simplify by eliminating common factors.

(X + 5) (X – 2) .  (2x+12)(x-3)
(3x-2)(x+5)        (x-2)(x-3)

This becomes:

(2x+12)
(3x-2).

Now we can further simplify:

2(x+6)
(3x-2)

This can also apply to equations.

1 + 4__      = 2
X   X + 2

Because this is an addition question, we must find a common denominator, we can do that by cross multiplying. Remember that this also applies to the other side of the equation.

1(X+2) + 4(X)___  = 2(X+2)(X)
X(X+2)    X(X + 2)       X(X + 2)

Now, we combine the fractions into one.

x + 2 + 4x  = 2X²+ 4x
X² + 2x              X² + 2x

Because both sides of the equation have a common denominator, we can remove the denominator entirely.

5x + 2 = 2x² + 4

From here, we are left with a quadratic equation.

-2x² + 5x + 2 = 4

2x² – 5x – 2 = -4

2x² – 5x + 2 = 0

(2x – 1) (x – 2) = 0

X₁ = ½

X­­­₂ = 2.

In this case, division and multiplication are the same, because solving a division question requires converting it to multiplication before anything else is done. Subtraction in quadratic expressions works the same way as addition, except with subtraction instead of addition.

This week in Pre calculus we learned about solving quadratic inequalities.

Solving quadratic inequalities is quite similar to solving quadratic equalities, but there are a few additional things that must be considered.

When solving quadratic inequalities, you must always keep in mind, that if you flip the positive/negatives of the inequality, you must also flip the sign, so if the sign is “less than”, you must flip it to “greater than”.

example:

0 < – x² + 6x + 9

First, we can flip the negative sign, and make it positive. But remember: when we do this, we must also flip the inequation symbol.

Which will look like this:

0 > x² + 6x + 9

Now, we can factor the inequation:

0 > (x + 3)²

Now, we must do the restrictions. We know that X cannot equal -3, because 0 is not greater than 0.

This week in Pre Calculus 11 we continued to learn about inequalities, specifically, graphing them. Inequalities are when two sides of an “equation” do not equate. Graphing inequalities is almost the same as graphing equalities, but with a few key differences. If the symbol is “≤” (less than or equal to) or “≥” (greater than or equal to), you draw a solid line on the equation, then either shade everything below the line or above the line, respectively. If the symbol is “<” or “>”, you draw a dotted line on the equation to indicate that it is not included, then shade everything below the line or above the line, respectively.

You can see here, because Y is less than or equal to the right side, the shaded area will be below the line, and the line will be solid, instead of dotted.

In the image on the left, Y is greater than the right side, therefore the line will be dotted, and the part above the line will be shaded.

In the image on the left, Y is less than the right side, therefore, the line will be dotted, and the shaded area will be below the line.

This week in Pre Calculus 11 we learned about several terms in graphing.

Parabolas are identical on each side of the axis of symmetry.

Axis of symmetry:

the line in the middle of the parabola.

Vertex:

the highest or lowest point of the parabola, whichever is finite.

Root:

Roots are another way to say X-intercepts (where the parabola crosses the X-axis).

this week in precalc 11 we learned about quadratic equations.

Where the lines of your graph intersect will be the solution to the equation, if they do not intersect, there is no solution.

example:

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solution: x = 12.5, y = 5.

example tw0:

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the lines never intersect therefore there is no solution.

This week in Pre Calculus we learned about radical equations, and fractions with radicals.

For Example:

the next step here is to multiply both parts of the fraction by the bottom in order to rationalize it.

next step:

15√5

_____

5

The solution is then to simplify the fraction

solution:

3√5

But what about when we have two radicals on the bottom? We multiply by the conjugate of the bottom.

for example:

5

_____

√6 + √2

next step:

5             (√6 – √2)

_____

√6 + √2 (√6 – √2)

=

5√6 – 5√2

_____

6 – 2

=

5√6 – 5√2

_____

4

if possible we can then simplify, but it is not possible in this example.