Solution to a System

The solution to a system is the point of intersection. Solution

1. True
2. False

'Solve the system' means find the point of intersection (POI). The solution to the system is the POI.

Solutions

Match the systems to their number of solutions. Solution

$\begin{align} & y = -2x + 1 \\ \\ & y = -2x + 1 \\ \\ \end{align}$

$\begin{align} & y = -5x + 3 \\ \\ & y = -5x + 1 \\ \\ \end{align}$

$\begin{align} & y = 3x + 1 \\ \\ & y = 5x - 2 \\ \\ \end{align}$

0 (no) solution

1 solution

2 solutions

Infinite solutions

A system has 0 (no) solution when the lines are parallel (same slope).
Infinite solutions occur when the lines overlap (collinear).
1 solution is 1 point of intersection.

Solving Linear Systems with Elimination

Solve the system using elimination.

Solution $y = 4 - x$
$5x - y = 8$

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x = y =

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$\begin{align} \cancel{y} & = 4 - x \\ \\ \underline{(+) \ \ 5x - \cancel{y}} & = \underline{8} \\ \\ 5x & = 12 - x \\ \\ 6x & = 12 \\ \\ x & = 2 \\ \\ \end{align}$ Sub 'x' into an equation to solve for 'y' $\begin{align} y & = 4 - x \\ \\ y & = 4 - (2) \\ \\ y & = 2 \\ \\ \end{align}$ The solution to the system is the point of intersection, POI: (2, 2)

Solution $x - 2y + 3 = 0$
$y + x = 3$

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x = y =

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$\begin{align} \cancel{x} - 2y + 3 & = 0 \\ \\ \underline{(-) \ \ y + \cancel{x}} & = \underline{3} \\ \\ -3y + 3 & = -3 \\ \\ -3y & = -6 \\ \\ y & = 2 \\ \\ \end{align}$ Then solve for x (not shown)... x = 1

Solution $5a - 2b = 5$
$3a + 2b = 19$

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a = b =

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$\begin{align} 5a - 2b & = 5 \\ \\ \underline{(+) \ \ 3a + 2b} & = \underline{19} \\ \\ 8a + 0b & = 24 \\ \\ 8a & = 24 \\ \\ a & = \dfrac{24}{8} \\ \\ a & = 3 \\ \\ \end{align}$ Then solve for b when a = 3 $\begin{align} 5a - 2b & = 5 \\ \\ 5(3) - 2b & = 5 \\ \\ -2b & = 5 - 15 \\ \\ -2b & = -10 \\ \\ b & = \dfrac{-10}{-2} \\ \\ b & = 5 \\ \\ \end{align}$

Line Segments

Explain the difference between a right bisector and an altitude. [3] Solution

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Pre - Length of a line formula

The distance between two points can be calculated as the hypotenuse using the Pythagorean theorem. Given the two points A(1, -5) and B(3, n)...

Use the Pythagorean theorem to determine the distance between the two points, in terms of n. Solution

Use a sketch to visualize/determine the change in x and the change in y...
∆ x = 3 - 1 = 2
∆ y = n - (-5) = n + 5

Calculate hypotenuse length (distance), c... $\begin{align} (a)^2 + (b)^2 & = (c)^2 \\ \\ (x)^2 + (y)^2 & = (c)^2 \\ \\ (2)^2 + (n + 5)^2 & = c^2 \\ \\ 4 + n^2 + 5n + 5n + 25 & = c^2 \\ \\ n^2 + 10n + 29 & = c^2 \\ \\ c & = \pm \sqrt{n^2 + 10n + 29} \\ \\ \end{align}$

Now, calculate the length of the line between these two points using the distance formula below, in terms of n. Solution $\begin{align} d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ \end{align}$

Make A(1, -5) point 1, and B(3, n) point 2.
x₁ = 1
y₁ = -5
x₂ = 3
y₂ = n
$\begin{align} d & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ & = \sqrt{(3 - 1)^2 + (n - (-5))^2} \\ \\ & = \sqrt{(2)^2 + (n + 5)^2} \\ \\ & = \sqrt{4 + (n + 5)^2} \\ \\ & = \sqrt{4 + n^2 + 5n + 5n + 25} \\ \\ & = \sqrt{n^2 + 10n + 29} \\ \\ \end{align}$ Takeaway: See that the distance (length) of a line equation is based on the same idea as the Pythagorean theorem. It is important to understand this before you jump into the distance (length) of a line equation, so you know where it comes from.

Pre - Length of a line formula basics

True or false? Solution $\begin{align} \sqrt{(5 - x)^2 + (3 - y)^2} = \sqrt{(x - 5)^2 + (y - 3)^2} \\ \\ \end{align}$

1. True
2. False

This is true because the square of the differences for example, (5 - x)² or (x - 5)², becomes the same positive magnitude...

Which of the following operations is/are correct? Solution
$\begin{align} I) \quad\quad\quad d & = \sqrt{(x - 3)^2} \\ \\ & = x - 3 \\ \\ \end{align}$ $\begin{align} II) \quad\quad\quad d & = \sqrt{(x + 1)^2 + (x - 2)^2} \\ \\ & = (x + 1) + (x - 2) \\ \\ \end{align}$

1. I only
2. II only
3. I and II
4. Neither

II is wrong because when there are more than one term underneath the square root, then the root and squares no longer cancel. The root only cancels the squares when there is one square term underneath (example I).

Perpendicular lines, solving systems, and length of a line segment.

A line connects points X(1, 2) and Y(5, 7). A third point Z(-2, 8) is located nearby.

Determine the equation of the line connecting XY. Solution

y =

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$\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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Solve the system of the two perpendicular lines, XY and the perpendicular line from Z, ie. determine the point of intersection (POI). Simplify fully. Solution

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$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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Calculate the shortest distance from point Z to the line XY. Solution

d =

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Applications of Circles

Built in the 1400's, the Duomo in Florence, Italy is still the largest masonry dome in the world. The sides of the arched Duomo lie on part of the perimeters of two circles shown below, where the intersection forms the cross section of the dome. Determine the height of the Duomo if the radius of each circle is 43 m and the base diameter of the Duomo is 45 m. Solution

Creative Commons: Sailko, 2006

h =

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Put the x and y axis through the center of the base of the Duomo. This makes the point of intersection (POI) at x = 0 in the form... (0, y) According to the origin (0, 0), determine the center of a circle to make an equation. See that (45 - 43) gives us the horizontal distance between a center of the circle and the other circles nearest point on the perimeter. Using the circle on the right side, the center of the circle is located at (20.5, 0) $\begin{align} x & = \dfrac{45}{2} - (45 - 43) \\ \\ & = 22.5 - 2 \\ \\ & = 20.5 \\ \\ \end{align}$ The circle on the right side has its center translated from the origin by (h, k), which corresponds to the points we determined (20.5, 0).. The equation of the circle is: $\begin{align} (x - h)^2 + (y - k)^2 & = r^2 \\ \\ (x - 20.5)^2 + (y - 0)^2 & = 43^2 \\ \\ \end{align}$ From where the origin was placed, you can see the POI is at x = 0, now solve for y using the equation of the circle... $\begin{align} (0 - 20.5)^2 + (y)^2 & = 43^2 \\ \\ y^2 & = 43^2 - (-20.5)^2 \\ \\ y & = \pm \sqrt{1428.75} \\ \\ y & = \pm 38\ m \\ \\ \end{align}$ The height of the Duomo is 38 m. (Only use the positive value for y).

Applications of Circles, Chords, Distance

A volcano in Iceland erupts with center C(10, 14), the ash cloud formed has radius 3 km and can be modeled with the circle equation below. A weather drone plane flies on route y = 0.8x + 9.

Determine if the weather drone will enter the ash cloud. Prove your reasoning by showing your work in your notes. Solution

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Determine the point(s) of intersection (POI), if any...
Expand the circle equation and substitute the linear equation into it... $\begin{align} (x - 10)^2 + (y - 14)^2 & = 3^2 \\ \\ x^2 - 20x + 100 + y^2 - 28y + 196 & = 9 \\ \\ x^2 - 20x + y^2 - 28y + 287 & = 0 \\ \\ \end{align}$ Sub the linear equation y = 0.8x + 9... $\begin{align} x^2 - 20x + (0.8x + 9)^2 - 28(0.8x + 9) + 287 & = 0 \\ \\ x^2 - 20x + (0.64x^2 + 14.4x + 81) - 22.4x - 252 + 287 & = 0 \\ \\ 1.64x^2 - 28x + 116 & = 0 \\ \\ \end{align}$ Use the quadratic equation, where a = 1.64, b = -28, c = 116... $\begin{align} x & = \dfrac{-(-28) ± \sqrt{(-28)^2 \ - \ 4(1.64)(116)}}{2(1.64)} \\ \\ & = \dfrac{28 ± \sqrt{23.04}}{3.28} \\ \\ & = 7.07,\ 10 \\ \\ \end{align}$ Since there are 2 POIs then drone flies through the ash cloud...

The weather drone manufacturers built the drone with a warranty to withstand flying through a maximum of 3km of ash conditions. Determine if the drone should divert its course. Solution

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In other words, find the coordinates of the POIs to calculate the distance.

First find the coordinates by substituting the 'x' values into the equation... $\begin{align} y & = 0.8x + 9 \\ \\ & = 0.8(7.073) + 9 \\ \\ & = 14.66 \\ \\ \\ \\ y & = 0.8x + 9 \\ \\ & = 0.8(10) + 9 \\ \\ & = 17 \\ \\ \end{align}$ The points are (7.07, 14.66) and (10, 17)

Calculate the distance (the length of the chord)... $\begin{align} d & = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2} & = \sqrt{\left(10 - 7.07\right)^2 + \left(17 - 14.66\right)^2} & = \sqrt{14.04} & = 3.75\ km \end{align}$ The drone would fly 3.75 km through the ash cloud, and therefore should divert its course to remain under warranty.

Line Segments and Centers

Right bisectors are used to determine the centroid of a triangle. Solution

1. True
2. False

• Right bisectors are used to determine the circumcenter of a triangle (and circle).
• Altitudes are used to determine the orthocenter of a triangle
• Midpoints through the opposite vertex are used to determine the centroid of a triangle

Orthocenters and circumcenters can be located outside of their triangles, while centroids cannot. Solution

1. True
2. False

This is true in obtuse triangles. Orthocenters and circumcenters can be located outside of their triangles. Centroids are never located outside of their triangles.

Triangles and Distance

Given the following points, determine if triangle ABC is either scalene, equilateral, or isosceles. Solution A(-5, 2)     B(4, 2)     C(2, -5)

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Determine the 3 side length distances...

• 3 equal side lengths = equilateral
• 2 equal side lengths = isosceles
• no equal side lengths = scalene

AB... $\begin{align} d & = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2} \\ \\ d & = \sqrt{\left(4 - (-5)\right)^2 + \left(2 - 2\right)^2} \\ \\ d & = \sqrt{\left(9\right)^2 + \left(0\right)^2} \\ \\ d & = 9 \\ \\ \end{align}$ AC... $\begin{align} d & = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2} \\ \\ d & = \sqrt{\left(2 - (-5)\right)^2 + \left(-5 - 2\right)^2} \\ \\ d & = \sqrt{\left(7\right)^2 + \left(-7\right)^2} \\ \\ d & = \sqrt{98} \\ \\ \end{align}$ BC... $\begin{align} d & = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2} \\ \\ d & = \sqrt{\left(2 - 4\right)^2 + \left(-5 - 2\right)^2} \\ \\ d & = \sqrt{\left(-2\right)^2 + \left(-7\right)^2} \\ \\ d & = \sqrt{53} \\ \\ \end{align}$ Since none of the side lengths are equal, the triangle is scalene.

Right Bisectors

A right bisector is located at the midpoint on a line, and is perpendicular to the slope of the line. Given the following points, determine the equation of the right bisector of QR. Solution P(3, -1)     Q(5, 5)     R(0, 3)

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$\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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A right bisector is at the midpoint, and is perpendicular to the side.
Determine the midpoint of QR... $\begin{align} & = \left( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \right) \\ \\ & = \left( \dfrac{5 + 0}{2} , \dfrac{5 + 3}{2} \right) \\ \\ & = \left( \dfrac{5}{2} , 4 \right) \\ \\ \end{align}$ Determine the ⊥ slope of the side QR... $\begin{align} m & = \dfrac{y_2 - y_1}{x_2 - x_1} \\ \\ m & = \dfrac{3 - 5}{0 - 5} \\ \\ m & = \dfrac{-2}{-5} \\ \\ m & = \dfrac{2}{5} \\ \\ ⊥m & = -\dfrac{5}{2} \\ \\ \end{align}$ Determine the equation using ⊥m and the midpoint on the line... $\begin{align} y & = mx + b \\ \\ 4 & = -\dfrac{5}{2}\left(\dfrac{5}{2}\right) + b \\ \\ 4 & = -\dfrac{25}{4} + b \\ \\ b & = \dfrac{41}{4} \\ \\ \end{align}$ The equation of the right bisector of QR is... $\begin{align} y & = mx + b \\ \\ y & = -\dfrac{5}{2}x + \dfrac{41}{4} \\ \\ \end{align}$

Circumcenter of a Triangle

Given the following points, determine the circumcenter of the triangle. (Hint: a circumcenter is where all the right bisectors intersect.)

Determine the equation of the right bisector of LM. Solution

y =

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With a quick sketch you will see that this triangle is obtuse rather than acute... An you will remember that with obtuse triangles, the circumcenter is located outside of it...

Circumcenters are right bisectors... Find the midpoint... $\begin{align} \text{Midpoint} _{LM} & = \left( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \right) \\ \\ \text{Midpoint} _{LM} & = \left( \dfrac{2 + 1}{2} , \dfrac{2 + 1}{2} \right) \\ \\ \text{Midpoint} _{LM} & = \left( \dfrac{3}{2} , \dfrac{3}{2} \right) \\ \\ \end{align}$ Find the ⊥m of LM... $\begin{align} m & = \dfrac{y_2 - y_1}{x_2 - x_1} \\ \\ m & = \dfrac{1 - 2}{1 - 2} \\ \\ m & = 1 \\ \\ ⊥m & = -1 \\ \\ \end{align}$ Equation of right bisector of LM... $\begin{align} y & = mx + b \\ \\ \dfrac{3}{2} & = -1\left(\dfrac{3}{2}\right) + b \\ \\ b & = 3 \\ \\ \\ \\ \\ \\ y & = -x + 3 \\ \\ \end{align}$

Determine the equation of the right bisector of MN. Solution

y =

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$\dfrac{}{\quad\quad}x \ - \ \dfrac{}{\quad\quad}$

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With a quick sketch you will see that this triangle is obtuse rather than acute... An you will remember that with obtuse triangles, the circumcenter is located outside of it...

Circumcenters are right bisectors... Find the midpoint... $\begin{align} \text{Midpoint} _{MN} & = \left( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \right) \\ \\ \text{Midpoint} _{MN} & = \left( \dfrac{1 + 0}{2} , \dfrac{1 + (-3)}{2} \right) \\ \\ \text{Midpoint} _{MN} & = \left( \dfrac{1}{2} , -1 \right) \\ \\ \end{align}$ Find the ⊥m of MN... $\begin{align} m & = \dfrac{y_2 - y_1}{x_2 - x_1} \\ \\ m & = \dfrac{-3 - 1}{0 - 1} \\ \\ m & = 4 \\ \\ ⊥m & = -\dfrac{1}{4} \\ \\ \end{align}$ Equation of right bisector of MN... $\begin{align} y & = mx + b \\ \\ -1 & = -\dfrac{1}{4}\left(\dfrac{1}{2}\right) + b \\ \\ b & = -\dfrac{7}{8} \\ \\ \\ \\ \\ \\ y & = -\dfrac{1}{4}x -\dfrac{7}{8} \\ \\ \end{align}$

Determine the coordinate of the circumcenter. Reduce fully. Solution

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$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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The circumcenter is the point of intersection, POI... Find the point of intersection of the right bisectors of LM and MN... $\begin{align} ① \ y & = -x + 3 \\ \\ ② \ y & = -\dfrac{1}{4}x -\dfrac{7}{8} \\ \\ \end{align}$ Substitute ① in ② $\begin{align} -x + 3 & = -\dfrac{1}{4}x -\dfrac{7}{8} \\ \\ -\dfrac{3}{4}x & = - \dfrac{31}{8} \\ \\ x & = \dfrac{31}{6} \\ \\ \end{align}$ Determine y... $\begin{align} y & = -\left(\dfrac{31}{6}\right) + 3 \\ \\ y & = -\dfrac{13}{6} \\ \\ \end{align}$ Therefore the circumcenter is located at $\left(\dfrac{31}{6}, -\dfrac{13}{6} \right)$

Parabolas: a Simple Sketch First

Sketch the parabola for the following...

Given the table of values for the equation, y = x² - 4x + 4

x	y
-2	16
-1	9
0	4
1	1
2	0
3	1
4	4
5	9
6	16

Given just the equation, y = -3x² + 2

Hint, make your own table of values...

x	y = -3(x)2 + 2
bbb	ccc
bbb	ccc
bbb	ccc
bbb	ccc

And then sketch...

Degree of Expressions, with Extended Types

Determine the degree of each of the following expressions.

$\dfrac{5}{x^2}$ Solution

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$\begin{align} & = \dfrac{5}{x^2} \\ \\ & = 5x^{-2} \\ \\ \end{align}$ Degree = -2

$\dfrac{3x^2}{2x^5y^3}$ Solution

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Remember the degree is the highest sum of the exponents of variables in one term. The highest is -2 (and not -3).

$\begin{align} & = \dfrac{3x^2}{2x^5y^3} \\ \\ & = \dfrac{3}{2}x^{(2-5)}y^{-3} \\ \\ & = \dfrac{3}{2}x^{-3}y^{-3} \\ \\ \end{align}$ Degree = -6

x^-2y^-1 + x^-1y^-1 Solution

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Remember the degree is the highest sum of the exponents of variables in one term. The highest is -2 (and not -3).

Degree = -2

$5\sqrt{x} + 4x^{-1}$ Solution

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$\begin{align} & = 5\sqrt{x} + 4x^{-1} \\ \\ & = 5x^{\tfrac{1}{2}} + 4x^{-1} \\ \\ \end{align}$ Degree = ½

Functions: Vertical Line Test: Coordinate Pair Sets

Determine if the following sets are functions:

(-2, 4) (-1, 3) (4, 2) (5, 2) Solution

1. Yes it is a function
2. No it is not a function

Yes this is a function because there are no same values of 'x' with two different values of 'y'.

(-2, 3) (-2, 3) (-1, 7) (2, 3) (5, 4) Solution

1. Yes it is a function
2. No it is not a function

Yes this is still a function because
(-2, 3) and (-2, 3) are the same point and do not count as 2 different y-values for the same x-value

(-2, 5) (1, 3) (1, 5) (4, 1) (5, 2) Solution

1. Yes it is a function
2. No it is not a function

No this is not a function because the two points, (1, 3) and (1, 5) have 2 different y-values: 3 & 5 for the same x-value: 1

Solve for a point on a function given the x-value

Solve:

$\quad y = \dfrac{3}{4}x + 5 \ $ when $\ x = 12$ Solution

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$\quad j = -3n^2 + 4n - 3 \ $ when $\ n = 2$ Solution

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$\quad a = -3b^2 + 4bc - 10c + 1 \ $ when $\ b = -1, c = 4$ Solution

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Review of the Basics: Product Rule

Write each with a single, positive exponent, showing your work without using a calculator.

$3^2 × 3^5$ Solution

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= 3⁷

$5^2 × 5^{-2}$ Solution

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= 5^{2 + (-2)}

= 5⁰

= 1

$(-10)^{-5} × (-10)^7$ Solution

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= (-10)^{-5 + 7}

= (-10)²

$\left(\dfrac{2}{3}\right)^4 × \left(\dfrac{2}{3}\right)^3$ Solution

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$\left(\dfrac{2}{3}\right)$

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$= \left(\dfrac{2}{3}\right)^7$

Review of the Basics: Quotient Rule

Simplify, showing your work without using a calculator.

$7^7 \div 7^3$ Solution

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= 7^{7 - 3}

= 7⁴

$(-2)^{-3} \div (-2)^8$ Solution

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= (-2)^-11

$(-5)^{6} \div (-5)^{-4}$ Solution

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= (-5)^{6 - (-4)}

= (-5)¹⁰

Exponents: Power of a Power

Write each with a single, positive exponent, or a reduced fraction, showing your work without using a calculator.

$\left(3^2\right)^4$ Solution

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= 3^{2 × 4}

= 3⁸

$\left(2^2\right)^3$ Solution

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= 2^{2 × 3}

= 2⁶

$\left(\left(2^2\right)^3\right)^2$ Solution

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2^{2 × 3 × 2}

= 2¹²

$\left(\cfrac{1}{2}\right)^3$ Solution

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$\begin{align} & = \cfrac{1^3}{2^3} \\ \\ & = \cfrac{1}{2^3} \\ \\ & = \cfrac{1}{8} \end{align}$

$\left(\cfrac{3}{2^2}\right)^3$ Solution

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$\begin{align} & = \cfrac{3^3}{2^{2 × 3}} \\ \\ & = \cfrac{3^3}{2^6} \\ \\ & = \cfrac{27}{64} \\ \\ \end{align}$

Exponents: Negative Powers

Simplify fully to an integer or fraction using positive exponents, showing your work without using a calculator.

$3^{-2}$ Solution

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$\begin{align} & = \dfrac{1}{3^2} \\ \\ & = \dfrac{1}{9} \end{align}$

$-4^{-3}$ Solution

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$\begin{align} & = -4^{-3} \\ \\ & = -\dfrac{1}{4^3} \\ \\ & = \dfrac{-1}{64} \\ \\ \end{align}$

$\left( \dfrac{3}{4} \right)^{-1}$ Solution

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$\quad = \left( \dfrac{4}{3} \right)^{+1} = \dfrac{4}{3}$

$\left( \dfrac{2}{1} \right)^{-2}$ Solution

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$\quad \left( \dfrac{1}{2} \right)^{2} = \dfrac{1^2}{2^2} = \dfrac{1}{4}$

$\left( \dfrac{2}{3} \right)^{-3}$ Solution

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$\quad \left( \dfrac{3}{2} \right)^{3} = \dfrac{3^3}{2^3} = \dfrac{27}{8}$

$\cfrac{2}{3^{-2}}$ Solution

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$\begin{align} & = \cfrac{2 × 3^{2}}{1} \\ \\ & = \cfrac{2 × 9}{1} \\ \\ & = 18 \\ \\ \end{align}$

$\left(9^{-7}\right)^0$ Solution

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$\quad \left(9^{-7 × 0}\right) = 9^0 = 1$

Exponents: Combinations

Write each with a single, positive exponent, or a reduced fraction, showing your work without using a calculator.

$2^4 \div 2^{-2} × 2^{-3}$ Solution

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= 2^{4 - (-2) + (-3)}

= 2^{4 + 2 - 3}

= 2³

$\left(3^2 × 3^{-1}\right)^2 \div 3$ Solution

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$\left( \dfrac{3^{-2}}{3^{-4}} \right)^{-2}$ Solution

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Exponents: Combinations

Write with a single, positive exponent, showing your work without using a calculator. Solution $\quad \cfrac{\left( \cfrac{2^{5}}{2^{-2}} \right)^{-2}}{\left( \cfrac{4^{2}}{2^{3}} \right)^{-1}}$

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$\begin{align} & = \cfrac{\left( 2^{5 - (-2)} \right)^{-2}}{\left( \cfrac{4^{2}}{2^{3}} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^7 \right)^{-2}}{\left( \cfrac{4^{2}}{2^{3}} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^7 \right)^{-2}}{\left( \cfrac{(2^2)^{2}}{2^{3}} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^{7(-2)} \right)}{\left( \cfrac{2^{2(2)}}{2^{3}} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^{-14} \right)}{\left( \cfrac{2^{4}}{2^{3}} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^{-14} \right)}{\left( 2^{4 - 3} \right)^{-1}} \\ \\ & = \cfrac{\left( 2^{-14} \right)}{\left( 2^1 \right)^{-1}} \\ \\ & = \cfrac{\left( 2^{-14} \right)}{\left( 2^{1(-1)} \right)} \\ \\ & = \cfrac{\left( 2^{-14} \right)}{\left( 2^{-1} \right)} \\ \\ & = 2^{-14 - (-1)} \\ \\ & = 2^{-13} \\ \\ & = \dfrac{1}{2^{13}} \\ \\ \end{align}$

Number Sense: Exponent Rules with Variables, Including Negative Exponents

Simplify, answers should have positive exponents only.

$\quad\dfrac{x^6y\,^{-2}}{x^3y}$ Solution

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$\quad -{\big(3x^{-4}\big)}^{-2}$ Solution

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Exponential Equations

Solve the following exponential equations, similar to the steps shown in the example below. Simplify fully.

$3^{2x - 3} = 27$ Solution

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$\begin{align} 3^{2x - 3} & = 27 \\ \\ 3^{2x - 3} & = 3^3 \\ \\ 2x - 3 & = 3 \\ \\ 2x & = 6 \\ \\ x & = 3 \\ \\ \end{align}$

$8^{x - 2} = 2^{3 - 6x}$ Solution

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$\begin{align} 8^{x - 2} & = 2^{3 - 6x} \\ \\ (2^3)^{x - 2} & = 2^{3 - 6x} \\ \\ \cancel{2}^{3(x - 2)} & = \cancel{2}^{3 - 6x} \\ \\ 3(x - 2) & = 3 - 6x \\ \\ 3x - 6 & = 3 - 6x \\ \\ 9x & = 9 \\ \\ x & = 1 \\ \\ \end{align}$

$25 × 5^{4x} = 5^{3x + 4}$ Solution

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$\begin{align} 25 × 5^{4x} & = 5^{3x + 4} \\ \\ 5^2 × 5^{4x} & = 5^{3x + 4} \\ \\ \cancel{5}^{2 + 4x} & = \cancel{5}^{3x + 4} \\ \\ 2 + 4x & = 3x + 4 \\ \\ x & = 2 \\ \\ \end{align}$

$\left(3^{x^2}\right)\left(3^{4x}\right) = 3^{-4}$ Solution

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$\begin{align} \left(3^{x^2}\right)\left(3^{4x}\right) & = 3^{-4} \\ \\ 3^{x^2 + 4x} & = 3^{-4} \\ \\ x^2 + 4x & = -4 \\ \\ x^2 + 4x + 4 & = 0 \\ \\ (x + 2)(x + 2) & = 0 \\ \\ x & = -2 \\ \\ \end{align}$

$\left(7^x\right)\left(7^{x - 5}\right) = 1$ Solution

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$\begin{align} \left(7^x\right)\left(7^{x - 5}\right) & = 1 \\ \\ 7^{x + (x - 5)} & = 7^0 \\ \\ 7^{2x - 5} & = 7^0 \\ \\ 2x - 5 & = 0 \\ \\ 2x & = 5 \\ \\ x & = \dfrac{5}{2} \\ \\ \end{align}$

$\left(\dfrac{1}{2^{-6}}\right) = \left(\dfrac{1}{2}\right)^x$ Solution

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$\begin{align} \left(\dfrac{1}{2^{-6}}\right) & = \left(\dfrac{1}{2}\right)^x \\ \\ \left(\dfrac{1}{2}\right)^{-6} & = \left(\dfrac{1}{2}\right)^x \\ \\ x & = -6 \\ \\ \end{align}$

$\left(\dfrac{1}{8}\right)^{x} = 8^{x - 4}$ Solution

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$\begin{align} \left(\dfrac{1}{8}\right)^{x} & = 8^{x - 4} \\ \\ \left(\dfrac{1}{2^3}\right)^{x} & = (2^3)^{x - 4} \\ \\ \left(2^{-3}\right)^{x} & = (2)^{3(x - 4)} \\ \\ -3x & = 3(x - 4) \\ \\ -3x & = 3x - 12 \\ \\ -6x & = - 12 \\ \\ x & = 2 \\ \\ \end{align}$

$\left(9\right)^{x - 1} = \left(\dfrac{1}{3}\right)^{2x}$ Solution

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$\begin{align} \left(9\right)^{x - 1} & = \left(\dfrac{1}{3}\right)^{2x} \\ \\ \left(3^2\right)^{x - 1} & = \left(3^{-1}\right)^{2x} \\ \\ 3^{2(x - 1)} & = 3^{-1(2x)} \\ \\ 2x - 2 & = -2x \\ \\ 4x & = 2 \\ \\ x & = \dfrac{1}{2} \\ \\ \end{align}$

Basic Elements of Quadratic Functions

The table below shows the flight path of a ball, with distance and height in meters.

Find the vertex Solution

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Remember back to the optimization you learned last year. The vertex is located at the maximum or minimum of the y-values. In this table, there is a maximum located at (4, 17).

Determine the equation of the axis of symmetry Solution

x =

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The equation of the axis of symmetry is located at the x-value of the maximum point, which is:

x = 4

Find the maximum height Solution

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m

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The maximum height is the highest y-value, which is 17.

Don't forget to write the units:

Therefore the maximum height occurs at 17m.

Verify that the equation h = -x² + 8x + 1 can be used to model the flight path of the ball. [1] Solution

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You verify an equation by substituting 2 or more points into the equation and check that L.S. equals R.S. $\begin{align} (h) & = -(x)^2 + 8(x) + 1 \\ \\ (8) & = -(1)^2 + 8(1) + 1 \\ \\ 8 & = -1 + 8 + 1 \\ \\ 8 & = 8 \\ \\ L.S. & = R.S. \\ \\ \end{align}$ $\begin{align} (h) & = -(x)^2 + 8(x) + 1 \\ \\ (17) & = -(4)^2 + 8(4) + 1 \\ \\ 17 & = -16 + 32 + 1 \\ \\ 17 & = 17 \\ \\ L.S. & = R.S. \\ \\ \end{align}$ Yes, the equation can be used to model the flight path of the ball.

Finite Differences (First and Second)

Use finite differences to determine if the following relations are linear, quadratic, or neither.

Solution

x	y
0	4
1	6
2	8
3	10
4	12

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Calculate the first difference by subtracting each adjacent y-value. 4 - 6 = -2
6 - 8 = -2
8 - 10 = -2
10 - 12 = -2
All first differences are the same, therefore the relation is linear.

Solution

x	y
0	-19
1	-12
2	-7
3	-4
4	-3

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-19 - (-12) = -7
-12 - (-7) = -5
-7 - (-4) = -3
-4 - (-3) = -1
The first differences are not the same, therefore not linear. Find second differences: -7 - (-5) = -2
-5 - (-3) = -2
-3 - (-1) = -2
Since the second differences are the same, then the relation is quadratic.

x	y
1	0
3	1
5	8
7	27
9	64

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0 - 1 = -1
1 - 8 = -7
8 - 27 = -19
27 - 64 = -37

Translations and Vertex Form

Given the general, vertex form of a quadratic.

Which of the following values determines a vertical compression by a certain factor? Solution

1. ±
2. a
3. h
4. k
5. none of the above

Vertical compression when: 0 < |a| < 1.

Vertical stretch when: |a| > 1.

Which of the following values determines a reflection in the x-axis? Solution

1. ±
2. a
3. h
4. k
5. none of the above

Reflection on x-axis when: a < 0... when 'a' is negative. $\begin{align} y = -a(x - h)^2 + k \\ \\ \end{align}$ Parabola opens up when z > 0... when 'a' is positive. $\begin{align} y = a(x - h)^2 + k \\ \\ \end{align}$

Which of the following values determines the horizontal shift left or right? Solution

1. ±
2. a
3. h
4. k
5. none of the above

Careful, this is the tricky one!

+h = shift right $\begin{align} y & = a(x - (+h))^2 + k \\ \\ y & = a(x - h)^2 + k \\ \\ \end{align}$ -h = shift left $\begin{align} y & = a(x - (-h))^2 + k \\ \\ y & = a(x + h)^2 + k \\ \\ \end{align}$

Transformations of Quadratic Functions

Match the type of transformations - shown with equations - to their respective graphs. Solution

${\color{Lime}y = 5x^2} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = \dfrac{1}{2}x^2}$

$\color{Lime}{y = (x - 2)^2} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = (x + 2)^2}$

$\color{Lime}{y = -(x)^2} \\ \\ \color{DimGray}{y = x^2}$

$\color{Lime}{y = x^2 + 3} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = x^2 - 3}$

Vertical stretch or compression by a factor of 'a'... ${\color{Lime}y = 5x^2} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = \dfrac{1}{2}x^2}$ Horizontal shift/translation by 'h' units left or right... $\color{Lime}{y = (x - 2)^2} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = (x + 2)^2}$ Vertical reflection... $\color{Lime}{y = -(x)^2} \\ \\ \color{DimGray}{y = x^2}$ Vertical shift/translation by 'k' units up or down... $\color{Lime}{y = x^2 + 3} \\ \\ \color{DimGray}{y = x^2} \\ \\ \color{Blue}{y = x^2 - 3}$

Transformations of Quadratic Functions

State the transformations on the following quadratic functions compared to the parent function, $x^2$.

$f(x) = \dfrac{1}{4}x^2$ Solution

• Vertical compression by a factor of ¼

$f(x) = x^2 + 8$ Solution

• Vertical translation (shift) up 8 units

$f(x) = -3x^2$ Solution

• Reflection across the x-axis
• Vertical stretch by a factor of 3

$f(x) = -(x - 9)^2 + 1$ Solution Video

• Reflection across the x-axis
• Horizontal translation (shift) 9 units right
• Vertical translation (shift) 1 unit up

$f(x) = -2(x + 1)^2 - 3$ Solution

• Reflection across the x-axis
• Horizontal translation (shift) 1 unit left
• Vertical translation (shift) 3 units down

$2f(3x)$ Solution Video

• Vertical stretch by a factor of 2
• Horizontal compression by a factor of ⅓

$f(x-2) + 4$ Solution Video

• Horizontal translation (shift) 2 units right
• Vertical translation (shift) 4 units up

$-f(-x)$ Solution Video

• Reflection across the x-axis -ƒ(x)
• Reflection across the y-axis ƒ(-x)

Two Other Transformations of Quadratic Functions

Based on the value 'n' determine some of the horizontal transformations below.

The transformation to y = a[n(x - h)]² + k with a horizontal reflection. Solution

1. y = ⅔[(x - 2)]² + 1
2. y = -2[⅓(x - 5)]² + 5
3. y = -1[1(x + 3)]² - 4
4. y = 2[-2(x + 6)]² - 1
5. y = ⅕[(x - 1)]² + 2

A horizontal reflection is due to ƒ(-x), or in an other form, y = a[-n(x - h)]² + k.

For example consider the transformation of $y = \sqrt{x}$ to $y = \sqrt{-x}$.

The transformation to y = a[n(x - h)]² + k with a horizontal stretch by a factor of 5. (careful here) Solution

1. y = -1[⅕(x - 2)]² + 4
2. y = ⅕[n(x + 3)]² - 1
3. y = 1[-2(x - ⅕)]² - 2
4. y = 2[-5(x - 3)]² - 5
5. y = 1[3(x + 6)]² + ⅕

The vertical stretch or compression is by a factor of $\dfrac{1}{n}$ !

So for a horizontal stretch by a factor of 5, the 'n' value must be ⅕ ... because $\dfrac{1}{⅕} = 5$.

Sketching Graphs of Functions

Sketch the graphs of the following functions. Label the x-intercepts, the axis of symmetry, and the vertex for each.

$f(x) = (x + 2)(x - 2)$ Solution

x-intercepts: x = -2, +2

Equation of axis of symmetry: x = 0

Vertex: sub the axis of symmetry (x = 0) into function: $\begin{align} f(x) & = (x + 2)(x - 2) \\ \\ f(0) & = (0 + 2)(0 - 2) \\ \\ f(0) & = (2)(-2) \\ \\ f(0) & = -4 \\ \\ y & = -4 \\ \\ \end{align}$ Vertex: (0, -4)

$y = -2(x - 2)(x - 4)$

The Three Ways to Find a Vertex

Find the vertex using each of the following methods:

By completing the square: Solution $y = 2x^2 + 12x - 10$

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$\begin{align} y & = 2x^2 + 12x - 10 \\ \\ & = 2(x^2 + 6x) - 10 \\ \\ & = 2(x^2 + 6x + 9 - 9) - 10 \\ \\ & = 2(x^2 + 6x + 9) - 10 + 2(-9) \\ \\ & = 2(x + 3)(x + 3) - 10 - 18 \\ \\ & = 2(x + 3)^2 - 28 \\ \\ \end{align}$ Vertex: (-3, -28)

Using the formula for the x-value of the vertex: Solution $y = -4.5x^2 - 1.8x + 3$

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a = -4.5, b = -1.8, c = 3. $\begin{align} x & = \frac{-b}{2a} \\ \\ & = \frac{-(-1.8)}{2(-4.5)} \\ \\ & = \frac{1.8}{-9} \\ \\ & = -0.2 \\ \\ \\ \\ \\ \\ f(x) & = -4.5x^2 - 1.8x + 3 \\ \\ f(-0.2) & = -4.5(-0.2)^2 - 1.8(-0.2) + 3 \\ \\ & = -4.5(0.04) + 0.36 + 3 \\ \\ & = -0.18 + 0.36 + 3 \\ \\ & = 3.18 \\ \\ \end{align}$ Vertex: (-0.2, 3.18)

Solve using symmetry: Solution $y = 2(x - 1)(x + 5)$

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Roots at x = 1, -5.
Axis of symmetry at x = -2 $\begin{align} f(x) & = 2(x - 1)(x + 5) \\ \\ f(-2) & = 2((-2) - 1)((-2) + 5) \\ \\ & = 2(-3)(3) \\ \\ & = 2(-9) \\ \\ & = -18 \\ \\ \end{align}$ Vertex: (-2, -18)

Solving Using Symmetry in the Quadratic Relations Unit

A parabola has vertex (-3, 7), and one x-intercept is -11. Find the other x-intercept and the y-intercept (reduce fully). Solution

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x-intercept: (   ,   )
y-intercept:━━━

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First find the other x-intercept with symmetry. $-11 \xrightarrow{+8} -3 \xrightarrow{+8} \text{______}$
x = +5
The x-intercept is at (5, 0) To find the y-intercept, determine the equation first... plug in the points (-3, 7) and (-11, 0) $\begin{align} y & = a(x - h)^2 + k \\ \\ 0 & = a(-11 + 3)^2 + 7 \\ \\ 0 & = a(-8)^2 + 7 \\ \\ 0 & = 64a + 7 \\ \\ a & = \dfrac{-7}{64} \\ \\ \end{align}$ Then set x = 0... $\begin{align} y & = \dfrac{-7}{64}(x + 3)^2 + 7 \\ \\ y & = \dfrac{-7}{64}(0 + 3)^2 + 7 \\ \\ y & = \dfrac{-7}{64}(9) + 7 \\ \\ y & = \dfrac{385}{64} \\ \\ \end{align}$

Review of Basics

Simplify by collecting like terms.

(2x² – 3x + 1) – (–x² – 3x – 5) Solution

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Distribute the negative and remove parentheses, then collect like terms. $\begin{align} & = (2x^2 – 3x + 1) \ – 1(–x^2 – 3x – 5) \\ \\ & = 2x^2 – 3x + 1 + x^2 + 3x + 5 \\ \\ & = 2x^2 + x^2 + \cancel{3x – 3x} + 1 + 5 \\ \\ & = 3x^2 + 6 \\ \\ \end{align}$

(5x² – y) + (6y – 2x²) – (y² + 7x²) Solution

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Distribute the negative and remove parentheses, then collect like terms. $\begin{align} & = (5x^2 – y) \ + \ (6y – 2x^2) \ – (y^2 + 7x^2) \\ \\ & = 5x^2 – y + 6y – 2x^2 – y^2 – 7x^2 \\ \\ & = 5x^2 – 2x^2 – 7x^2 – y^2 – y + 6y \\ \\ & = -4x^2 – y^2 + 5y \\ \\ \end{align}$

Care with Distributive Property of Terms

Find the term that has the mistake in the following statement. Solution 3x(2x - 3) = 6x - 9x

1. 3x
2. 2x
3. 6x
4. 9x
5. None of the above

6x is wrong because is an incorrect use of the distributive property. It should be 6x².

3x(2x - 3) = 6x² - 9x

FOIL (A + B)(C + D) = AC + AD + BC + BD

Expand and simplify. Write in standard form.

$\quad (x + 3)(x + 4)$ Solution

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$\quad -(4x - 1)(3x + 2)$ Solution

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FOIL first: $\begin{align} & = -(4x - 1)(3x + 2) \\ \\ & = -(12x^2 + 8x - 3x - 2) \\ \\ & = -(12x^2 + 5x - 2) \\ \\ \end{align}$ Then distribute the negative sign into each term in parenthesis: $\begin{align} & = -1(12x^2 + 5x - 2) \\ \\ & = -12x^2 - 5x + 2 \\ \\ \end{align}$

$\quad (8x - 4)^2$ Solution

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! Careful, do not do this: $(8^2x^2 - 4^2)$

FOIL first, then collect like terms: $\begin{align} & = (8x - 4)(8x - 4) \\ \\ & = 64x^2 - 32x - 32x + 16 \\ \\ & = 64x^2 - 64x + 16 \\ \\ \end{align}$

$\quad 5(x + 2)(3x + 1) - 2(x + 2)$ Solution

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$\quad 7(2x - 3)^2$ Solution

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$\quad (2x - 5y)(6x - 6y)$ Solution

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FOIL first, then collect like terms. $\begin{align} & = (2x)(6x) + (2x)(-6y) + (-5y)(6x) + (-5y)(-6y) \\ \\ & = 12x^2 -12xy -30xy + 30y^2 \\ \\ & = 12x^2 -42xy + 30y^2 \\ \\ \end{align}$

Expanding Using FOIL

Determine the expanded standard form of (2x - 3)(4x + 1), hence FOIL. Solution

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FOIL and put your answer in standard form

$\begin{align} & = (2x - 3)(4x + 1) \\ \\ & = (2x)(4x) + (2x)(1) + (-3)(4x) + (-3)(1) \\ \\ & = 8x^2 + 2x - 12x -3 \\ \\ & = 8x^2 - 10x -3 \\ \\ \end{align}$

Expanding using FOIL is the reverse of factoring (and factoring is the reverse of FOIL).

1. True
2. False

FOIL with roots

Expand and simplify. $\quad (\sqrt{6}\,x + \sqrt{3})(\sqrt{6}\,x - \sqrt{3})$ Solution

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FOIL and put your answer in standard form

FOIL first, then collect like terms. $\begin{align} & = (\sqrt{6}\,x + \sqrt{3})(\sqrt{6}\,x - \sqrt{3}) \\ \\ & = (\sqrt{6}\,x)(\sqrt{6}\,x) + (\sqrt{6}\,x)(- \sqrt{3}) + (\sqrt{3})(\sqrt{6}\,x) + (\sqrt{3})(- \sqrt{3}) \\ \\ & = 6x^2 - \sqrt{6×3}\,x + \sqrt{6×3}\,x - 3 \\ \\ & = 6x^2 - \sqrt{18}\,x + \sqrt{18}\,x - 3 \\ \\ & = 6x^2 \cancel{- \sqrt{18}\,x + \sqrt{18}\,x} - 3 \\ \\ & = 6x^2 - 3 \\ \\ \end{align}$

Factoring Quadratic Expressions, Common Factoring

The greatest common factor (GCF) is used to common factor an expression. Determine the GCF of the following. Solution Video 16x²y + 8xy + 12y²

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Determine the GCF of the coefficients and the variables.

4 is the greatest factor of the numbers 16, 8, and 12.

The highest degree of the common exponent is y¹, or just y. $\begin{align} & = 16x^2y + 8xy + 12y^2 \\ \\ & = 4(4x^2y + 2xy + 3y^2) \\ \\ & = 4y(4x^2 + 2x + 3y) \\ \\ \end{align}$ GCF = 4y

Factoring Quadratic Expressions, Common Factoring

Factor.

$\quad 4x^2 + 12x$ Solution

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$\begin{align} & = 4x^2 + 12x \\ \\ & = 4(x^2 + 3x) \\ \\ & = 4x(x + 3) \\ \\ \end{align}$

$\quad 8x^2y + 12xy$ Solution

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$\begin{align} & = 8x^2y + 12xy \\ \\ & = 4(2x^2y + 3xy) \\ \\ & = 4x(2xy + 3y) \\ \\ & = 4xy(2x + 3) \\ \\ \end{align}$

$\quad 25yx^2 - 15xy^2 - 10xy$ Solution Video

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The highest number you can factor out is 5.
It doesn't matter if it's xy or yx...

The highest degree of x that can be common factored is x¹.
The highest degree of y that can be common factored is y¹.

Shown in steps: $\begin{align} & = 25yx^2 - 15xy^2 - 10xy \\ \\ & = 5(5yx^2 - 3xy^2 - 2xy) \\ \\ & = 5x(5yx - 3y^2 - 2y) \\ \\ & = 5xy(5x - 3y - 2) \\ \\ \end{align}$

Factor by Grouping

Factor the following, with the grouping already done for you. (Grouping step is an optional intermediate step). Solution 2x(3x - 1) - 2(3x - 1)

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This shows the step midway through factoring by grouping:

2x(3x - 1) - 2(3x - 1)

= (2x - 2)(3x - 1)

Factoring Quadratic Expressions, Trinomials when a = 1

Factor, using any method.

$\quad x^2 + 7x + 12$ Solution

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For factoring quadratics where there's nothing but 1 in-front of the x²...
The numbers that multiply to 12:
1 × 12, 2 × 6, 3 × 4
Choose the pair that adds to 7... 3 × 4
= (x + 3)(x + 4)

Or you could do factor by grouping: $\begin{align} & = x^2 + \underline{7x} + 12 \\ \\ & = x^2 + \underline{3x + 4x} + 12 \\ \\ & = x(x+3) + 4(x+3) \\ \\ & = (x+3)(x+4) \\ \\ \end{align}$

$\quad x^2 + 19x - 20$ Solution

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The numbers that multiply to -20:
-1 × 20, 1 × -20, -2 × 10, 2 × -10, -4 × 5, 4 × -5
The pair that adds to +19: -1 × 20
= (x - 1)(x + 20)

Or you could do factor by grouping: $\begin{align} & = x^2 + \underline{19x} - 20 \\ \\ & = x^2 + \underline{-1x + 20x} - 20 \\ \\ & = x(x-1) + 20(x-1) \\ \\ & = (x - 1)(x + 20) \\ \\ \end{align}$

$\quad x^2 - 5x - 24$ Solution

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The numbers that multiply to -24:
-1 × 24, 1 × -24, -2 × 12, 2 × -12, -3 × 8, 3 × -8, -4 × 6, 4 × 6
= (x + 3)(x - 8)

Or you could do factor by grouping: $\begin{align} & = x^2 - \underline{5x} - 24 \\ \\ & = x^2 +\underline{3x -8x} -24 \\ \\ & = x(x+3)-8(x+3) \\ \\ & = (x + 3)(x - 8) \\ \\ \end{align}$

Factoring Quadratic Expressions, Factor by Grouping to Prepare for Trinomials with a ≠ 1

Factor the following Solution 3x² + 2x - 8

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Which of the following is factored correctly? Solution -30x² + 130x + 100

1. y = -10(3x - 10)(x - 1)
2. y = -10(3x + 2)(x - 5)
3. y = -10(3x - 10)(x - 1)
4. y = -10(x - 10)(3x - 1)

Common factor the greatest common multiple first... $\begin{align} y & = -30x^2 + 130x + 100 \\ \\ y & = -10(3x^2 - 13x - 10) \\ \\ y & = -10(3x + 2)(x - 5) \\ \\ \end{align}$

Faster Factoring Without Decomposition

Now that you understand more about factoring it's time to take the training wheels off - no more decomposition (aka splitting and grouping). Factor straight into the parenthesis like the example shown below.

x² - 11x + 18 Solution

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What multiplies to +18 and adds to -11?
= -9, -2... $\begin{align} & = (\underbrace { x - \underbrace { 9)(x }_{ -9 } -2} _{ -2 } ) \\ \\ \end{align}$

3x² - 7x - 6 Solution

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What multiplies to -18 and adds to -7?
= -9, +2... $\begin{align} & = (\underbrace { 3x+\underbrace { 2)(x }_{ +2 } -3 }_{ -9 } ) \\ \\ & = (3x + 2)(x - 3) \\ \\ \end{align}$

9x² + 7x - 2 Solution

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What multiplies to -18 and adds to +7?
= +9, -2... $\begin{align} & = (\underbrace { 9x \ -\underbrace { 2)(x }_{ -2 } +1 }_{ +9 } ) \\ \\ & = (9x - 2)(x + 1) \\ \\ \end{align}$

8x² + 2x - 3 Solution

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$\begin{align} & = (\underbrace { \overbrace { 4x+\underbrace { 3)(2x }_{ +6 } }^{ +8 } -1 }_{ -4 } ) \\ \\ \end{align}$

-10x² - 22x - 4 Solution

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Common factor out the negative first, and common factor the $\begin{align} & = -(10x^2 + 22x + 4) \\ \\ & = -(\underbrace { \overbrace { 5x+\underbrace { 1)(2x }_{ +2 } }^{ +10 } +4 }_{ +20 } ) \\ \\ \end{align}$ It is considered better form to factor the 2 out, $\begin{align} & = -2(5x^2 + 11x + 2) \\ \\ & = -2(\underbrace { 5x+\underbrace { 1)(x }_{ +1 } +2 }_{ +10 } ) \\ \\ \end{align}$

Factoring Quadratic Expressions, Perfect Squares

Factor the following perfect squares fully, using any method (direct is encouraged). The general formulas for perfect squares are given below.

x² + 2x + 1 Solution

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(x + 1)²

x² + 10x + 25 Solution

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(x + 5)²

x² - 8x + 16 Solution

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(x - 4)²

x² - 20x + 100 Solution

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(x - 10)²

4x² - 12x + 9 Solution

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(2x - 3)² or (3 - 2x)²

Quadratic Expressions That Are Not Factorable

Explain in your own words why the following two examples of quadratics are not factorable. Solution y = x² + 6x + 12
y = x² - 3x + 10

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They are not factorable when no two numbers add to the middle term and multiply to the product of the first and last terms.

(Also accept a more in-depth exploration of this).

Factoring Quadratic Expressions, Perfect Squares

Given the general formula for a perfect square quadratic...

Determine the value of n that would make the following trinomial a perfect square. Solution 9x² + nx + 16

n =

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Find the value of n that would make the following trinomial a perfect square, and then write it as a perfect square. Solution 25x² + 30x + n²

n =

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$\begin{align} & 25x^2 + 30x + n^2 \\ \\ & = 5^2x^2 + 2(5)(n)x + \underline{n}^2 \\ \\ \end{align}$ See that 30x = 2(5)(n)x and solve... $\begin{align} & = 5^2x^2 + 2(5)(3)x + (\underline{3})^2 \\ \\ & = (5x + 3)(5x + 3) \\ \\ & = (5x + 3)^2 \\ \\ \end{align}$

Tough Quadratic

Factor fully: $\quad x^4 - 41x^2 + 400$ Solution

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Determine the factor pairs that multiply to +400 and add to -41...
±1 & ±400
±2 & ±200
±4 & ±100
±5 & ±80
±8 & ±50
±10 & ±40
-16 & -25 !

Factor by splitting and grouping (shown here)...
= x⁴ - 25x² - 16x² + 400
= x²(x² - 25) - 16(x² - 25)
= (x² - 16)(x² - 25) <-- Recognize the difference of squares!
= (x + 4)(x - 4)(x + 5)(x - 5)

Algebraic Expressions and Factoring

The dimensions for two rectangles are given below. Determine an expression for the total area of the two rectangles combined, then combine like terms and factor this expression fully. Solution

	Rectangle A	Rectangle B
Length (cm)	(2x - 3)	4
Width (cm)	3x	(x - 1)

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Add the areas, $\begin{align} & = (2x - 3)(3x) + 4(x - 1) \\ \\ & = 6x^2 - 9x + 4x - 4 \\ \\ & = 6x^2 - 5x - 4 \end{align}$ Then factor the expression, $\begin{align} & = 6x^2 - 5x - 4 \\ \\ & = 6x^2 + 3x - 8x - 4 \\ \\ & = 3x(2x + 1) - 4(2x + 1) \\ \\ & = (3x - 4)(2x + 1) \end{align}$

Algebraic Expressions and Factoring

A triangular sandbox is surrounded by a rectangular field of grass. The dimensions of the sandbox and field are listed below, in terms of x.

Determine a standard form expression for the area of the grass surrounding the sandbox. Solution

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Sandbox: x² + 4x

Field: 4x² + 0x - 4

Combined Expression: 3x² - 4x - 4 $\begin{align} A_{grass} & = A_{rectangle} + A_{triangle} \\ \\ & = (length)(width) + \dfrac{1}{2}(base)(height) \\ \\ & = (x + 1)(4x - 4) + \dfrac{1}{2}(2x)(x + 4) \\ \\ & = 4x^2 - 4x + 4x - 4 - \dfrac{1}{2}(2x^2 + 8x) \\ \\ & = 4x^2 - 4 - x^2 - 4x \\ \\ & = 3x^2 - 4x - 4 \\ \\ \end{align}$

Factor the expression fully. Solution

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3x² - 4x - 4

= 3x² + 2x - 6x - 4

= x(3x + 2) - 2(3x + 2)

= (3x + 2)(x - 2)

Solving Factored Quadratics Using Zero Product Property

Given the equation already in factored form.

How many solutions will there be? Solution

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solutions

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One solution will come from the first, x = 0
The other solution will come from the second, 2x - 3 = 0

Therefore there are 2 solutions.

Solve using the zero-product property. Solution

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x₁ = x₂ = ━━

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Remember that if $\begin{align} (a)(b) = 0 \\ \\ \end{align}$ Then a = 0, or b = 0...

So for, $\begin{align} x(2x - 3) = 0 \\ \\ \end{align}$ There are two answers:
$\begin{align} x = 0 \\ \\ \end{align}$ $\begin{align} 2x - 3 = 0 \\ \\ 2x = 3 \\ \\ x = \dfrac{3}{2} \\ \\ \end{align}$