Unit 1 · Quadratic Equations
1
Quadratic Formula & Discriminant
Unit 1 · Quadratics
Quadratic Formula: For $ax^2+bx+c=0$, $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Discriminant $\Delta = b^2-4ac$:
· $\Delta > 0$ → two distinct real roots
· $\Delta = 0$ → one repeated real root
· $\Delta < 0$ → no real roots (two complex roots)
Discriminant $\Delta = b^2-4ac$:
· $\Delta > 0$ → two distinct real roots
· $\Delta = 0$ → one repeated real root
· $\Delta < 0$ → no real roots (two complex roots)
📌 Worked Example
Solve $x^2-5x+6=0$$\Delta = 25-24 = 1 > 0$ → two real roots
$x = \frac{5 \pm 1}{2}$ → $x=3$ or $x=2$ ✓
Q1
Quadratic Formula
What are the solutions of $2x^2 - 7x + 3 = 0$?
✦ Solution
Apply the quadratic formula with $a=2,\, b=-7,\, c=3$.$\Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25$
$x = \dfrac{7 \pm 5}{4}$ → $x = \dfrac{12}{4} = 3$ or $x = \dfrac{2}{4} = \dfrac{1}{2}$
Answer: B
Q2
Vertex Form
Find the vertex of the parabola $f(x) = 2x^2 - 8x + 5$.
✦ Solution
Vertex x-coordinate: $x = -\dfrac{b}{2a} = -\dfrac{-8}{2(2)} = \dfrac{8}{4} = 2$Substitute: $f(2) = 2(4) - 8(2) + 5 = 8 - 16 + 5 = -3$
Vertex $= (2,\,-3)$ Answer: C
Q3
Discriminant
How many real roots does $x^2 + 4x + 7 = 0$ have?
✦ Solution
$\Delta = b^2 - 4ac = (4)^2 - 4(1)(7) = 16 - 28 = -12$Since $\Delta < 0$, there are no real roots (two complex conjugate roots).
Answer: A
Unit 2 · Polynomials
2
Polynomial Roots & Factor Theorem
Unit 2 · Polynomials
Factor Theorem: $(x-r)$ is a factor of $p(x)$ if and only if $p(r)=0$.
Rational Root Theorem: Possible rational roots of $a_nx^n+\cdots+a_0$ are $\pm\dfrac{\text{factors of }a_0}{\text{factors of }a_n}$.
Rational Root Theorem: Possible rational roots of $a_nx^n+\cdots+a_0$ are $\pm\dfrac{\text{factors of }a_0}{\text{factors of }a_n}$.
📌 Worked Example
Is $(x-2)$ a factor of $p(x)=x^3-3x^2+4$?$p(2)=8-12+4=0$ ✓ → Yes, $(x-2)$ is a factor.
Q4
Polynomial Roots
Which of the following gives all real roots of $x^3 - 2x^2 - 5x + 6 = 0$?
✦ Solution
Test $x=3$: $27-18-15+6=0$ ✓ → $(x-3)$ is a factor.Divide: $x^3-2x^2-5x+6 = (x-3)(x^2+x-2) = (x-3)(x+2)(x-1)$
Roots: $x = 3,\;-2,\;1$ Answer: B
Unit 3 · Exponential & Logarithmic Functions
3
Exponents & Logarithms
Unit 3 · Exp & Log
Key Rules:
· $\log_b(xy) = \log_b x + \log_b y$
· $\log_b\!\left(\dfrac{x}{y}\right) = \log_b x - \log_b y$
· $\log_b(x^n) = n\log_b x$
· $b^{\log_b x} = x$ (inverse property)
· $\log_b(xy) = \log_b x + \log_b y$
· $\log_b\!\left(\dfrac{x}{y}\right) = \log_b x - \log_b y$
· $\log_b(x^n) = n\log_b x$
· $b^{\log_b x} = x$ (inverse property)
📌 Worked Example
Solve $2^x = 64$: $2^x = 2^6$ → $x = 6$Expand: $\log\!\left(\dfrac{x^2}{y}\right) = 2\log x - \log y$
Q5
Logarithm Evaluation
What is the value of $\log_2 32$?
✦ Solution
$\log_2 32 = x$ means $2^x = 32 = 2^5$Therefore $x = 5$ Answer: B
Q6
Exponential Equation
Solve for $x$: $\;3^{2x} = 81$
✦ Solution
$3^{2x} = 81 = 3^4$Equate exponents: $2x = 4$ → $x = 2$ Answer: B
Q7
Logarithm Properties
Which expression is equivalent to $\log\!\left(\dfrac{x^2}{y}\right)$?
✦ Solution
Apply quotient rule: $\log\!\left(\dfrac{x^2}{y}\right) = \log x^2 - \log y$Then power rule: $= 2\log x - \log y$ Answer: B
Unit 4 · Sequences & Series
4
Arithmetic & Geometric Sequences
Unit 4 · Sequences & Series
Arithmetic: $a_n = a_1 + (n-1)d$ · $S_n = \dfrac{n}{2}(2a_1 + (n-1)d)$
Geometric: $a_n = a_1 \cdot r^{n-1}$ · $S_n = a_1 \cdot \dfrac{r^n - 1}{r-1}$ (if $r \neq 1$)
Geometric: $a_n = a_1 \cdot r^{n-1}$ · $S_n = a_1 \cdot \dfrac{r^n - 1}{r-1}$ (if $r \neq 1$)
📌 Worked Example
Arithmetic: $a_1=2,\;d=3$ → $a_5 = 2+4(3)=14$Geometric: $a_1=3,\;r=2$ → $a_4 = 3 \cdot 2^3 = 24$
Q8
Geometric Sequence
In a geometric sequence with $a_1 = 4$ and common ratio $r = 3$, what is $a_5$?
✦ Solution
$a_5 = a_1 \cdot r^{5-1} = 4 \cdot 3^4 = 4 \cdot 81 = 324$ Answer: C
Q9
Arithmetic Series Sum
Find the sum of the first 10 terms of an arithmetic sequence with $a_1=3$ and $d=4$.
✦ Solution
$S_{10} = \dfrac{10}{2}\bigl(2(3)+(10-1)(4)\bigr) = 5(6+36) = 5 \times 42 = 210$ Answer: C
Unit 5 · Complex Numbers
5
Complex Number Operations
Unit 5 · Complex Numbers
$i = \sqrt{-1}$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$
Multiplication: Use FOIL and replace $i^2$ with $-1$.
Conjugate: $(a+bi)(a-bi) = a^2 + b^2$ (always real)
Multiplication: Use FOIL and replace $i^2$ with $-1$.
Conjugate: $(a+bi)(a-bi) = a^2 + b^2$ (always real)
📌 Worked Example
$(2+3i)(1-i) = 2-2i+3i-3i^2 = 2+i+3 = 5+i$
Q10
Complex Multiplication
Simplify $(3+2i)(1-4i)$.
✦ Solution
$(3+2i)(1-4i) = 3 - 12i + 2i - 8i^2$$= 3 - 10i - 8(-1) = 3 - 10i + 8 = 11 - 10i$ Answer: C
Unit 6 · Absolute Value & Radical Equations
6
Absolute Value & Radical Equations
Unit 6 · Equations
$|A| = k$ (where $k \geq 0$): $A = k$ or $A = -k$
Radical equation: Isolate the radical, then square both sides.
⚠️ Always check for extraneous solutions after squaring.
Radical equation: Isolate the radical, then square both sides.
⚠️ Always check for extraneous solutions after squaring.
📌 Worked Example
$|x-1|=4$ → $x-1=4$ giving $x=5$, or $x-1=-4$ giving $x=-3$$\sqrt{x+1}=3$ → $x+1=9$ → $x=8$ ✓ (check: $\sqrt{9}=3$)
Q11
Absolute Value Equation
Solve: $|2x - 3| = 7$
✦ Solution
Case 1: $2x-3=7$ → $2x=10$ → $x=5$Case 2: $2x-3=-7$ → $2x=-4$ → $x=-2$
Check: $|2(5)-3|=7$ ✓ $|2(-2)-3|=|-7|=7$ ✓ Answer: B
Q12
Radical Equation
Solve: $\sqrt{2x + 3} = 5$
✦ Solution
Square both sides: $2x+3 = 25$$2x = 22$ → $x = 11$
Check: $\sqrt{2(11)+3} = \sqrt{25} = 5$ ✓ Answer: B
Unit 7 · Systems & Rational Equations
7
Systems of Equations & Rational Equations
Unit 7 · Systems
Solving systems: Use substitution or elimination.
Rational equations: Multiply both sides by the LCD. Check for excluded values (denominators ≠ 0).
Rational equations: Multiply both sides by the LCD. Check for excluded values (denominators ≠ 0).
📌 Worked Example
System: $x+y=5,\;x-y=1$ → add: $2x=6$ → $x=3,\;y=2$Rational: $\dfrac{1}{x} = 3$ → $x = \dfrac{1}{3}$
Q13
System of Equations
Solve the system: $\;2x + y = 7\;$ and $\;x - y = 2$
✦ Solution
Add the two equations: $(2x+y)+(x-y)=7+2$ → $3x=9$ → $x=3$Substitute: $2(3)+y=7$ → $y=1$
Solution: $(3,\,1)$ Answer: B
Q14
Rational Equation
Solve: $\;\dfrac{x+2}{x-1} = 3$
✦ Solution
Multiply both sides by $(x-1)$: $x+2 = 3(x-1) = 3x-3$$5 = 2x$ → $x = \dfrac{5}{2}$
Check: $x \neq 1$ ✓ Answer: C
Unit 8 · Conic Sections & Inverse Functions
8
Conic Sections & Inverse Functions
Unit 8 · Conics & Inverses
Circle: $(x-h)^2+(y-k)^2=r^2$, center $(h,k)$, radius $r$
Inverse: To find $f^{-1}(x)$, swap $x$ and $y$, then solve for $y$.
$(f \circ f^{-1})(x) = x$ always.
Inverse: To find $f^{-1}(x)$, swap $x$ and $y$, then solve for $y$.
$(f \circ f^{-1})(x) = x$ always.
📌 Worked Example
Circle center $(-1, 2)$, $r=4$: $(x+1)^2+(y-2)^2=16$Inverse of $f(x)=3x-9$: swap → $x=3y-9$ → $f^{-1}(x)=\dfrac{x+9}{3}$
Q15
Circle Equation
Write the standard equation of a circle with center $(2,\,-3)$ and radius $5$.
✦ Solution
Standard form: $(x-h)^2+(y-k)^2=r^2$Center $(2,-3)$: $h=2,\;k=-3$, $r=5$ so $r^2=25$
$(x-2)^2+(y-(-3))^2=25$ → $(x-2)^2+(y+3)^2=25$ Answer: C
Q16
Inverse Functions
If $f(x) = 2x + 6$, what is $f^{-1}(x)$?
✦ Solution
Set $y=2x+6$. Swap $x$ and $y$: $x=2y+6$Solve for $y$: $x-6=2y$ → $y=\dfrac{x-6}{2}$
Check: $f\!\left(\dfrac{x-6}{2}\right)=2\cdot\dfrac{x-6}{2}+6=x$ ✓ Answer: B
Unit 9 · Binomial Theorem & Exponential Growth
9
Binomial Theorem & Exponential Models
Unit 9 · Advanced Topics
Binomial Theorem: $(x+y)^n = \displaystyle\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k$
Exponential Growth: $P(t)=P_0 \cdot e^{rt}$, where $r$ = growth rate, $t$ = time
Exponential Growth: $P(t)=P_0 \cdot e^{rt}$, where $r$ = growth rate, $t$ = time
📌 Worked Example
Coeff. of $x^2$ in $(x+3)^4$: $\binom{4}{2}(3)^2=6 \times 9=54$$P(5) = 500 e^{0.1 \times 5} = 500e^{0.5} \approx 824.4$
Q17
Binomial Theorem
What is the coefficient of $x^3$ in the expansion of $(x+2)^5$?
✦ Solution
The term with $x^3$: $\binom{5}{2} x^3 \cdot 2^2$ (since $k=2$, $n-k=3$)$= 10 \cdot x^3 \cdot 4 = 40x^3$
Coefficient $= 40$ Answer: B
Q18
Exponential Growth
A population of 1000 grows continuously at 5% per year. Approximately how many are there after 10 years? ($e^{0.5} \approx 1.649$)
✦ Solution
$P(10) = 1000 \cdot e^{0.05 \times 10} = 1000 \cdot e^{0.5}$$= 1000 \times 1.649 = 1649$ Answer: B
Unit 10 · Probability & Statistics
10
Probability & Standard Deviation
Unit 10 · Probability & Stats
Addition Rule: $P(A \cup B) = P(A)+P(B)-P(A \cap B)$
Standard Deviation: $\sigma = \sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n}}$
Measures how spread out data values are from the mean.
Standard Deviation: $\sigma = \sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n}}$
Measures how spread out data values are from the mean.
📌 Worked Example
$P(A)=0.3,\;P(B)=0.5,\;P(A \cap B)=0.1$$P(A \cup B) = 0.3+0.5-0.1 = 0.7$
Q19
Probability
Events $A$ and $B$ satisfy $P(A)=0.4$, $P(B)=0.5$, $P(A \cap B)=0.2$. Find $P(A \cup B)$.
✦ Solution
$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$= 0.4 + 0.5 - 0.2 = 0.7$ Answer: B
Q20
Standard Deviation
The data set is $\{2,\,4,\,4,\,4,\,5,\,5,\,7,\,9\}$. The mean is $5$. What is the population standard deviation?
✦ Solution
Deviations from mean (5): $-3,-1,-1,-1,0,0,2,4$Squared deviations: $9,1,1,1,0,0,4,16$
Variance $= \dfrac{9+1+1+1+0+0+4+16}{8} = \dfrac{32}{8} = 4$
$\sigma = \sqrt{4} = 2$ Answer: B
✦ Final Results
–/ 20
–
Correct
–
Wrong
–
Time Used
✦ Answer Key & Solutions Summary
Q1
B — x = 3 or ½
Quadratic Formula
Q2
C — (2, −3)
Vertex of Parabola
Q3
A — No real roots
Discriminant
Q4
B — x = −2, 1, 3
Polynomial Roots
Q5
B — 5
Logarithm
Q6
B — x = 2
Exponential Equation
Q7
B — 2 log x − log y
Log Properties
Q8
C — 324
Geometric Sequence
Q9
C — 210
Arithmetic Series
Q10
C — 11 − 10i
Complex Numbers
Q11
B — x = 5 or x = −2
Absolute Value
Q12
B — x = 11
Radical Equation
Q13
B — (3, 1)
System of Equations
Q14
C — x = 5/2
Rational Equation
Q15
C — (x−2)²+(y+3)²=25
Circle Equation
Q16
B — (x−6)/2
Inverse Function
Q17
B — 40
Binomial Theorem
Q18
B — 1649
Exponential Growth
Q19
B — 0.7
Probability
Q20
B — 2
Standard Deviation