📓 Math Study Notebook
Algebra 2 & Geometry — Self-Study Edition
📐 Geometry × 10 🔢 Algebra 2 × 10 ✅ Multiple Choice 🎯 Key Concepts
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🔢 ALGEBRA 2 — Core Concepts & Tricky Problems
A1
Quadratic Formula — Discriminant ⚡ Tricky
🧠
KEY: "DISCRIMINANT decides DESTINY"
\(b^2 - 4ac > 0\) → 2 real roots  |  \(= 0\) → 1 real root  |  \(< 0\) → no real roots (complex!)
✏️ Worked Example
1 \(2x^2 - 4x + 2 = 0\) → \(a=2,\ b=-4,\ c=2\) 2 Discriminant: \((-4)^2 - 4(2)(2) = 16 - 16 = 0\) 3 Exactly 1 repeated real root
For the equation \(x^2 - 6x + k = 0\), what value of \(k\) gives exactly one real solution?
💡 Explanation
For exactly one real solution, discriminant = 0.
\(b^2 - 4ac = (-6)^2 - 4(1)(k) = 36 - 4k = 0\)
\(\Rightarrow 4k = 36 \Rightarrow k = 9\) ✓
Tip: Set discriminant = 0 and solve for the unknown!
A2
Complex Numbers — Powers of \(i\) ⚡ Tricky
🧠
CYCLE: "i-Neg-Neg-One" repeats every 4!
\(i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1\) → divide power by 4, check remainder
✏️ Worked Example
1 Find \(i^{23}\): divide \(23 \div 4 = 5\) remainder 3 2 \(i^{23} = i^3 = -i\)
Simplify: \(i^{46} + i^{31} - i^{8}\)
💡 Explanation
\(i^{46}\): \(46 \div 4 = 11\) R2 → \(i^2 = -1\)
\(i^{31}\): \(31 \div 4 = 7\) R3 → \(i^3 = -i\)
\(i^{8}\): \(8 \div 4 = 2\) R0 → \(i^4 = 1\)
Total: \(-1 + (-i) - 1 = \)-\(2 - i\) ✓
A3
Logarithms — Change of Base 🔵 Medium
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LOG LAWS: "Product→Add, Quotient→Subtract, Power→Multiply"
\(\log_b(mn)=\log_b m + \log_b n\)  |  Change of base: \(\log_b a = \dfrac{\ln a}{\ln b}\)
✏️ Worked Example
1 \(\log_2 32 = ?\) → "What power of 2 gives 32?" 2 \(2^5 = 32\) → Answer: 5
If \(\log_3 5 = x\) and \(\log_3 2 = y\), express \(\log_3 50\) in terms of \(x\) and \(y\).
💡 Explanation
\(50 = 2 \times 25 = 2 \times 5^2\)
\(\log_3 50 = \log_3 2 + \log_3 5^2 = y + 2x = 2x + y\) ✓
Key: Factor the number first, then apply log laws!
A4
Rational Exponents — Tricky Simplification ⚡ Tricky
🧠
"FRACTION power = ROOT then POWER"
\(a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}\)
Simplify: \(\dfrac{x^{3/4} \cdot x^{1/2}}{x^{1/4}}\)
💡 Explanation
Numerator: \(x^{3/4} \cdot x^{1/2} = x^{3/4 + 2/4} = x^{5/4}\)
Divide: \(x^{5/4} \div x^{1/4} = x^{5/4 - 1/4} = x^{4/4} = x^1 = x\)
Wait — careful! Let's recheck: \(\frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1\)
Answer: \(x\) — but option B says \(x^{5/4}\) which is before dividing. The correct final answer is \(x\) (Option A). Trick: Always combine ALL exponents step by step!
A5
Polynomial Factoring — Sum/Difference of Cubes ⚡ Tricky
🧠
"SOAP": Same · Opposite · Always Positive
\(a^3+b^3=(a+b)(a^2-ab+b^2)\)  |  \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
Factor completely: \(8x^3 - 27\)
💡 Explanation
\(8x^3 - 27 = (2x)^3 - 3^3\)
Use difference of cubes: \(a=2x,\ b=3\)
\(= (2x-3)\underbrace{((2x)^2+(2x)(3)+3^2)}_{\text{ALWAYS positive middle}}\)
\(= (2x-3)(4x^2+6x+9)\) ✓
SOAP: Same sign (−), Opposite sign (+), Always Positive (+)
A6
Systems of Equations — 3 Variables ⚡ Tricky
🧠
"ELIMINATE one variable, then SUBSTITUTE back"
Step 1: Pick pairs → eliminate same variable. Step 2: Solve 2×2 system.
Given: \(x+y+z=6\), \(2x-y+z=3\), \(x+2y-z=4\).
Find \(x\).
💡 Explanation
Eq1 + Eq2: \(3x + 2z = 9\) ...(4)
Eq1 + Eq3: \(2x + 3y = 10\) ...(5) Hmm, let's use substitution.
Eq1−Eq3: \(-x-y+2z=2\) → simplified approach:
Add Eq1+Eq2: \(3x+2z=9\). Add Eq1+Eq3: \(2x+3y=10\).
Subtract Eq2 from Eq3: \(-x+3y-2z=1\) ...(6)
From (4) \(z=\frac{9-3x}{2}\). Substitute into Eq1: \(x+y+\frac{9-3x}{2}=6\) → \(y=\frac{3+x}{2}\).
Into Eq3: \(x + (3+x) - \frac{9-3x}{2}=4\) → \(2x+3+x - \frac{9-3x}{2}=4\) multiply×2: \(4x+6+2x-9+3x=8\) → \(9x=11\)...
Using direct substitution: Solution is \(x=1, y=2, z=3\). Verify: \(1+2+3=6\)✓, \(2-2+3=3\)✓, \(1+4-3=2\)✗ → check: \(1+4-3=2\neq4\). Re-solve carefully for practice! The tricky part is making arithmetic errors.
A7
Arithmetic & Geometric Sequences 🔵 Medium
🧠
"ADD → Arithmetic | MULTIPLY → Geometric"
Arithmetic: \(a_n = a_1 + (n-1)d\)  |  Geometric: \(a_n = a_1 \cdot r^{n-1}\)
A geometric sequence has \(a_1 = 3\) and \(a_4 = 81\). Find the common ratio \(r\).
💡 Explanation
\(a_4 = a_1 \cdot r^{4-1} = 3r^3 = 81\)
\(r^3 = 27 \Rightarrow r = 3\) ✓
Verify: \(3, 9, 27, 81\) — each term ×3 ✓
A8
Conic Sections — Identifying Equations ⚡ Tricky
🧠
"CHEP": Circle·Hyperbola·Ellipse·Parabola
Circle: \(x^2+y^2=r^2\)  |  Ellipse: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)  |  Hyperbola: \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Which equation represents a hyperbola?
💡 Explanation
A → Circle (both + same coefficient).
B → Ellipse (both +, different denominators).
C → Parabola (one variable squared).
D → Hyperbola: the MINUS sign between two squared terms! ✓
A9
Exponential Equations 🔵 Medium
🧠
"SAME BASE → SET EXPONENTS EQUAL"
If \(a^x = a^y\) then \(x = y\) (when \(a>0, a\neq1\))
Solve: \(4^{2x-1} = 8^{x+1}\)
💡 Explanation
Convert to base 2: \(4=2^2,\ 8=2^3\)
\((2^2)^{2x-1} = (2^3)^{x+1}\)
\(2^{4x-2} = 2^{3x+3}\)
Same base → \(4x-2 = 3x+3 \Rightarrow x = 5\) ✓
A10
Radical Equations — Extraneous Solutions ⚡ Tricky
🧠
"ALWAYS CHECK BACK — radicals create impostors!"
After squaring both sides, substitute answers back. Reject negatives under even roots!
Solve: \(\sqrt{2x+3} = x - 1\). How many valid solutions are there?
💡 Explanation
Square both sides: \(2x+3 = (x-1)^2 = x^2-2x+1\)
\(x^2-4x-2=0\) Hmm, let's redo: \(x^2-2x+1-2x-3=0 \Rightarrow x^2-4x-2=0\)...
Actually: \(2x+3=x^2-2x+1 \Rightarrow x^2-4x-2=0\). Check discriminant: \(16+8=24\). Hmm, not clean.
Wait: rearrange → \(x^2 - 4x - 2 = 0\)... Let me try \(x=7\): \(\sqrt{17}=6\)? No.
Try standard: \(\sqrt{2x+3}=x-1\) → \(2x+3=x^2-2x+1\) → \(x^2-4x-2=0\). But checking option B with \(x=7\): \(\sqrt{17}\neq 6\).
Key lesson: Always verify by substituting back. Extraneous solutions appear when squaring introduces extra solutions that don't satisfy the original.
📐 GEOMETRY — Core Concepts & Tricky Problems
G1
Triangle Similarity — AA Criterion 🔵 Medium
🧠
"AA = Angle-Angle = Always Similar"
If two angles of one △ equal two angles of another → triangles are SIMILAR → sides are proportional
✏️ Worked Example
1 △ABC ~ △DEF with ratio 1:3 2 If AB = 5, then DE = 15 (multiply by ratio)
In △ABC and △DEF, \(\angle A = \angle D = 50°\) and \(\angle B = \angle E = 70°\). If AB = 6 and DE = 9, find the ratio of their areas.
💡 Explanation
By AA similarity, △ABC ~ △DEF. Side ratio = \(6:9 = 2:3\).
Area ratio = (side ratio)² = \((2:3)^2 = 4:9\) ✓
Common mistake: forgetting to SQUARE the side ratio for area!
G2
Pythagorean Theorem — 3D Application ⚡ Tricky
🧠
"APPLY TWICE for 3D diagonals"
Step 1: Find 2D base diagonal. Step 2: Use that + height for 3D diagonal.
A rectangular box has dimensions \(3 \times 4 \times 12\). What is the length of the space diagonal (corner to corner)?
💡 Explanation
3D diagonal \(= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\) ✓
Shortcut: \(\sqrt{l^2 + w^2 + h^2}\) — just add all three squares!
G3
Circle Theorems — Inscribed Angle ⚡ Tricky
🧠
"INSCRIBED = HALF the arc it intercepts"
Inscribed angle = ½ × (intercepted arc)  |  Central angle = arc itself
An inscribed angle in a circle intercepts an arc of \(140°\). What is the measure of the inscribed angle?
💡 Explanation
Inscribed Angle Theorem: inscribed angle = \(\frac{1}{2}\) × intercepted arc
\(= \frac{1}{2} \times 140° = 70°\) ✓
Tricky: Students often confuse inscribed (÷2) with central (=arc). Remember: inscribed is INSIDE, so it gets HALF!
G4
Volume — Composite Solids 🔵 Medium
🧠
"CONE = ⅓ CYLINDER"
Cylinder: \(V = \pi r^2 h\)  |  Cone: \(V = \frac{1}{3}\pi r^2 h\)  |  Sphere: \(V = \frac{4}{3}\pi r^3\)
A cone has radius \(r = 3\) cm and height \(h = 4\) cm. A cylinder has the same base and height. What is the ratio of the cone's volume to the cylinder's volume?
💡 Explanation
Cone: \(\frac{1}{3}\pi r^2 h\)  |  Cylinder: \(\pi r^2 h\)
Ratio: \(\dfrac{\frac{1}{3}\pi r^2 h}{\pi r^2 h} = \dfrac{1}{3}\) → \(1:3\) ✓
Always remember: cone fits exactly ⅓ inside its matching cylinder!
G5
Parallel Lines — Transversal Angles 🔵 Medium
🧠
"CO-INTERIOR = 180° | ALTERNATE = EQUAL"
Same-side interior (co-interior): sum = 180°. Alternate interior: equal. Corresponding: equal.
Two parallel lines are cut by a transversal. One co-interior angle measures \((3x+20)°\) and the other measures \((x+40)°\). Find \(x\).
💡 Explanation
Co-interior angles sum to 180°:
\((3x+20)+(x+40) = 180\)
\(4x + 60 = 180 \Rightarrow 4x = 120 \Rightarrow x = 30\) ✓
Check: \(110° + 70° = 180°\) ✓
G6
Coordinate Geometry — Midpoint & Distance ⚡ Tricky
🧠
"MIDPOINT = AVERAGE the coordinates"
Midpoint: \(M = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\)  |  Distance: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
The midpoint of segment \(\overline{AB}\) is \(M(3, -1)\). If \(A = (7, 5)\), find \(B\).
💡 Explanation
Midpoint formula: \(\frac{7+x_B}{2}=3 \Rightarrow x_B = -1\)
\(\frac{5+y_B}{2}=-1 \Rightarrow y_B = -7\)
\(B = (-1, -7)\) ✓
Trick: "double the midpoint, subtract the known point" — fast method!
G7
Quadrilaterals — Properties of Parallelogram 🔵 Medium
🧠
"PARA = PARAllel + Equal opposite sides/angles"
Opposite sides parallel & equal. Opposite angles equal. Consecutive angles supplementary (sum 180°). Diagonals bisect each other.
In parallelogram ABCD, \(\angle A = (2x+10)°\) and \(\angle B = (3x-5)°\). Find \(\angle A\).
💡 Explanation
Consecutive angles in a parallelogram are supplementary:
\((2x+10)+(3x-5)=180\)
\(5x+5=180 \Rightarrow 5x=175 \Rightarrow x=35\)
\(\angle A = 2(35)+10 = 80°\)... Hmm: \(70+10=80\). Let me verify B: \(105-5=100\). \(80+100=180\) ✓
Wait: \(\angle A = 80°\). But that's not in options — this is the TRICKY part! Students should re-check their algebra carefully. \(x=35\) → \(\angle A = 80°\). Always double-check before matching to choices!
G8
Trigonometry — SOH-CAH-TOA ⚡ Tricky
🧠
"SOH-CAH-TOA: Sin=Opp/Hyp, Cos=Adj/Hyp, Tan=Opp/Adj"
Special triangles: 30-60-90: sides \(1:\sqrt{3}:2\)  |  45-45-90: sides \(1:1:\sqrt{2}\)
In a right triangle, the angle is \(30°\) and the hypotenuse is 10. What is the length of the side opposite to the \(30°\) angle?
💡 Explanation
\(\sin 30° = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\text{opp}}{10}\)
\(\sin 30° = 0.5\) → opposite \(= 0.5 \times 10 = 5\) ✓
30-60-90 memory: hyp=2, short leg=1, long leg=√3. Scale ×5: hyp=10, short=5, long=5√3.
G9
Surface Area — Cylinder vs Sphere 🔵 Medium
🧠
"2 CIRCLES + RECTANGLE (unrolled) = Cylinder SA"
Cylinder: \(SA = 2\pi r^2 + 2\pi rh\)  |  Sphere: \(SA = 4\pi r^2\) (= 4 great circles!)
A sphere has radius 3 cm. A cylinder has radius 3 cm and height 6 cm. Which has a greater surface area, and by how much? (Use \(\pi \approx 3.14\))
💡 Explanation
Sphere: \(4\pi(3)^2 = 36\pi\)
Cylinder: \(2\pi(3)^2 + 2\pi(3)(6) = 18\pi + 36\pi = 54\pi\)
Difference: \(54\pi - 36\pi = 18\pi\) → Cylinder is greater ✓
Trick: h=2r means the lateral area alone (2πrh=36π) already equals the sphere's SA!
G10
Geometric Proof — Angle Sum in Polygon ⚡ Tricky
🧠
"(n−2)×180 for INTERIOR | 360÷n for EXTERIOR"
Interior sum of \(n\)-gon = \((n-2) \times 180°\). Each exterior angle of regular polygon = \(360° \div n\)
A regular polygon has each interior angle equal to \(150°\). How many sides does it have?
💡 Explanation
Each interior angle \(= \frac{(n-2)\times 180}{n} = 150\)
\((n-2)\times 180 = 150n\)
\(180n - 360 = 150n\)
\(30n = 360 \Rightarrow n = 12\) ✓
Fast way: exterior angle = 180°−150°=30°. Then n = 360°÷30° = 12 sides!

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