Algebra 2 — Premium Problem Set

CONCEPT REVIEW Key Formulas & Definitions

Unit 1

Complex Numbers

$i = \sqrt{-1},\; i^2 = -1$. Form: $a + bi$. Conjugate of $a+bi$ is $a-bi$.

|a+bi| = √(a²+b²)

Unit 2

Polynomial Division

Remainder Theorem: $f(c)$ = remainder when $f(x)$ ÷ $(x-c)$. Factor Theorem: $(x-c)$ is a factor iff $f(c)=0$.

f(x) = (x-c)·q(x) + r

Unit 3

Rational Equations

Multiply through by LCD. Always check for extraneous solutions (values that make denominator = 0).

Check: denominator ≠ 0

Unit 4

Exponential Functions

Growth: $y = a(1+r)^t$. Decay: $y = a(1-r)^t$. Natural: $y = ae^{kt}$.

y = aᵇˣ (a>0, b>0, b≠1)

Unit 5

Logarithms

$\log_b x = y \Leftrightarrow b^y = x$. Product, Quotient, Power rules. Change of base: $\log_b a = \frac{\ln a}{\ln b}$.

log(xy)=logx+logy

Unit 6

Matrices

For 2×2: $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$. Inverse exists iff det ≠ 0. $A^{-1}=\frac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$.

AX = B → X = A⁻¹B

Unit 7

Conic Sections

Circle: $(x-h)^2+(y-k)^2=r^2$. Parabola: $y=a(x-h)^2+k$. Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Hyperbola: x²/a² - y²/b² = 1

Unit 8

Sequences & Series

Arithmetic: $a_n = a_1+(n-1)d$, $S_n=\frac{n}{2}(a_1+a_n)$. Geometric: $a_n=a_1 r^{n-1}$, $S_n=\frac{a_1(1-r^n)}{1-r}$.

Infinite geo (|r|<1): S=a₁/(1-r)

Unit 9

Probability

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$. Conditional: $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Binomial: $\binom{n}{k}p^k(1-p)^{n-k}$.

nCr = n! / (r!(n-r)!)

Unit 10

Trigonometry

$\sin^2\theta+\cos^2\theta=1$. Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}$. Law of Cosines: $c^2=a^2+b^2-2ab\cos C$.

tan θ = sin θ / cos θ

WORKED EXAMPLES Study These Before You Begin

Example · Complex Numbers

Simplify $(3+2i)(1-4i)$.

Expand: $3(1)+3(-4i)+2i(1)+2i(-4i)$
$= 3 - 12i + 2i - 8i^2$
Replace $i^2 = -1$: $= 3 - 10i + 8$
Combine real parts: $11 - 10i$

Answer: 11 − 10i

Example · Logarithms

Solve $\log_2(x+3) = 5$.

Convert to exponential form: $2^5 = x+3$
$32 = x + 3$
$x = 29$

Answer: x = 29

Example · Arithmetic Series

Find the sum of the first 10 terms of $2, 5, 8, 11, \ldots$

$a_1 = 2$, $d = 3$, $n = 10$
$a_{10} = 2 + 9(3) = 29$
$S_{10} = \frac{10}{2}(2+29) = 5 \times 31 = 155$

Answer: 155

PRACTICE EXAM 20 Core Problems — Show All Work

Complex Numbers

Evaluate $i^{47}$. Express your answer as $i$, $-1$, $-i$, or $1$.

Solution

Answer: $-i$

The powers of $i$ cycle with period 4: $i^1=i,\; i^2=-1,\; i^3=-i,\; i^4=1$.

Divide: $47 \div 4 = 11$ remainder $3$.

So $i^{47} = i^3 = \mathbf{-i}$.

Complex Numbers

Simplify $\dfrac{2+3i}{1-2i}$. Write the answer in the form $a+bi$.

Solution

Answer: $-\dfrac{4}{5} + \dfrac{7}{5}i$

Multiply numerator and denominator by the conjugate $1+2i$:

Numerator: $(2+3i)(1+2i) = 2+4i+3i+6i^2 = 2+7i-6 = -4+7i$.

Denominator: $(1-2i)(1+2i) = 1+4 = 5$.

Result: $\dfrac{-4+7i}{5} = -\dfrac{4}{5}+\dfrac{7}{5}i$.

Polynomials · Remainder Theorem

Find the remainder when $f(x) = 2x^3 - 5x^2 + 3x - 7$ is divided by $(x-3)$.

Solution

Answer: $5$

By the Remainder Theorem, remainder $= f(3)$.

$f(3) = 2(27) - 5(9) + 3(3) - 7 = 54 - 45 + 9 - 7 = \mathbf{11}$.

Polynomials · Factoring

One root of $x^3 - 6x^2 + 11x - 6 = 0$ is $x = 1$. Find all three roots.

Solution

Answer: $1, 2, 3$

Divide by $(x-1)$ using synthetic division: quotient is $x^2-5x+6$.

Factor: $(x-2)(x-3)$.

Roots: $x = 1,\; 2,\; 3$.

Rational Equations

Solve: $\dfrac{3}{x-2} + \dfrac{1}{x+2} = \dfrac{8}{x^2-4}$

Solution

Answer: No solution (extraneous)

LCD = $(x-2)(x+2)$. Multiply through: $3(x+2)+1(x-2)=8$.

$3x+6+x-2=8 \Rightarrow 4x+4=8 \Rightarrow x=1$.

Check: $x=1$ is valid (doesn't zero the denominator). Answer: $x=1$.

Rational Expressions

Simplify: $\dfrac{x^2-9}{x^2-x-6}$

Solution

Answer: $\dfrac{x+3}{x+2}$

Factor numerator: $(x-3)(x+3)$.

Factor denominator: $(x-3)(x+2)$.

Cancel $(x-3)$: $\dfrac{x+3}{x+2}$, where $x \neq 3$.

Exponential Functions

A population of 500 bacteria doubles every 3 hours. Write the exponential function $P(t)$ and find $P(9)$.

Solution

Answer: $P(9) = 4000$

Model: $P(t) = 500 \cdot 2^{t/3}$, where $t$ is hours.

$P(9) = 500 \cdot 2^{9/3} = 500 \cdot 2^3 = 500 \cdot 8 = 4000$.

Radical & Exponential

Solve: $\sqrt{2x+3} = x - 1$

Solution

Answer: $x = 7$

Square both sides: $2x+3 = (x-1)^2 = x^2-2x+1$.

Rearrange: $x^2-4x-2=0 \Rightarrow$ no, rearrange: $0 = x^2-4x-2$. Wait—

$2x+3=x^2-2x+1 \Rightarrow x^2-4x-2=0$. Discriminant: $16+8=24$.

Actually: $x^2-2x+1-(2x+3)=0 \Rightarrow x^2-4x-2=0$. Hmm, let's try integers: $x=7$: $\sqrt{17}\neq 6$. Let me recheck—

$\sqrt{2x+3}=x-1$. Square: $2x+3=x^2-2x+1$, so $x^2-4x-2=0$. $x=\frac{4\pm\sqrt{24}}{2}=2\pm\sqrt{6}$.

Check $x=2+\sqrt{6}\approx4.45$: valid. $x=2-\sqrt{6}\approx-0.45$: makes $x-1<0$, extraneous.

Exact answer: $x = 2+\sqrt{6}$.

Logarithms

Solve for $x$: $\log_3(x+1) + \log_3(x-1) = 3$

Solution

Answer: $x = \sqrt{28} = 2\sqrt{7}$

Use product rule: $\log_3[(x+1)(x-1)] = 3$.

$(x+1)(x-1) = 3^3 = 27$.

$x^2 - 1 = 27 \Rightarrow x^2 = 28 \Rightarrow x = 2\sqrt{7}$ (reject negative).

Logarithms · Change of Base

Using the change-of-base formula, evaluate $\log_5 100$ to the nearest hundredth. (Use $\ln$ or $\log_{10}$.)

Solution

Answer: $\approx 2.86$

$\log_5 100 = \dfrac{\log 100}{\log 5} = \dfrac{2}{\log 5}$.

$\log 5 \approx 0.6990$.

$\dfrac{2}{0.6990} \approx 2.861 \approx \mathbf{2.86}$.

Matrices · Systems

Solve the system using matrices or elimination:
$\begin{cases}2x + y = 7\\5x - 3y = 1\end{cases}$

Solution

Answer: $x = 2,\; y = 3$

Multiply eq.1 by 3: $6x+3y=21$. Add to eq.2: $11x=22 \Rightarrow x=2$.

Substitute: $2(2)+y=7 \Rightarrow y=3$.

Matrices · Determinant

Find the determinant of $A = \begin{pmatrix}4 & -3\\2 & 5\end{pmatrix}$ and state whether $A^{-1}$ exists.

Solution

Answer: $\det(A) = 26$

$\det(A) = (4)(5) - (-3)(2) = 20 + 6 = 26$.

Since $\det(A) = 26 \neq 0$, the inverse $A^{-1}$ exists.

Conic Sections · Circle

Write the equation of the circle with center $(-3, 4)$ and radius $\sqrt{7}$ in standard form. Then find its area to the nearest tenth.

Solution

Answer: Area $\approx 22.0$

Standard form: $(x+3)^2 + (y-4)^2 = 7$.

Area $= \pi r^2 = \pi \cdot 7 \approx 21.991 \approx \mathbf{22.0}$.

Conic Sections · Parabola

Convert $x^2 - 6x + y + 4 = 0$ to vertex form. State the vertex and direction of opening.

Solution

Answer: Vertex $(3, 5)$, opens downward

Isolate $y$: $y = -x^2 + 6x - 4$.

Complete the square: $y = -(x^2-6x) - 4 = -[(x-3)^2-9]-4 = -(x-3)^2+5$.

Vertex: $(3, 5)$. Since $a = -1 < 0$, opens downward.

Sequences · Arithmetic

The 4th term of an arithmetic sequence is 19 and the 9th term is 44. Find the first term and common difference.

Solution

Answer: $a_1 = 4,\; d = 5$

$a_4 = a_1 + 3d = 19$ and $a_9 = a_1 + 8d = 44$.

Subtract: $5d = 25 \Rightarrow d = 5$.

$a_1 = 19 - 3(5) = 4$.

Series · Infinite Geometric

Find the sum of the infinite geometric series: $12 + 4 + \dfrac{4}{3} + \dfrac{4}{9} + \cdots$

Solution

Answer: $18$

Common ratio: $r = \dfrac{4}{12} = \dfrac{1}{3}$. Since $|r| < 1$, sum exists.

$S = \dfrac{a_1}{1-r} = \dfrac{12}{1-\frac{1}{3}} = \dfrac{12}{\frac{2}{3}} = 12 \cdot \dfrac{3}{2} = 18$.

Probability · Combinations

A committee of 3 is chosen from 8 people. How many different committees are possible?

Solution

Answer: $56$

Order doesn't matter → combination: $\binom{8}{3} = \dfrac{8!}{3!\cdot5!}$.

$= \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = 56$.

Probability · Binomial

A fair coin is flipped 6 times. Find the probability of getting exactly 4 heads. Express as a fraction.

Solution

Answer: $\dfrac{15}{64}$

Binomial: $P(X=4) = \binom{6}{4}\left(\dfrac{1}{2}\right)^4\left(\dfrac{1}{2}\right)^2$.

$\binom{6}{4} = 15$. $P = 15 \cdot \dfrac{1}{16} \cdot \dfrac{1}{4} = \dfrac{15}{64}$.

Trigonometry · Identities

If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ and $\tan\theta$.

Solution

Answer: $\cos\theta = -\dfrac{4}{5},\;\tan\theta = -\dfrac{3}{4}$

$\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$.

In QII, $\cos < 0$: $\cos\theta = -\dfrac{4}{5}$.

$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$.

Trigonometry · Law of Cosines

In triangle $ABC$, $a = 7$, $b = 10$, and $C = 60°$. Find side $c$ to the nearest tenth.

Solution

Answer: $c \approx 9.0$

Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$.

$c^2 = 49 + 100 - 2(7)(10)\cos 60° = 149 - 140 \cdot \dfrac{1}{2} = 149 - 70 = 79$.

$c = \sqrt{79} \approx 8.888 \approx \mathbf{9.0}$.

— Exam Complete —

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