Self-Study Guide
Core Problems
20 essential problems across Algebra 2 & Geometry — most-missed topics, with memory keys and full explanations.
Part I
Algebra 2
Quadratic Equations & Functions
01
⚡ DISCRIMINANT = b²−4ac
How many real solutions does 2x² − 4x + 5 = 0 have?TRAP
Quick Check Method
Discriminant D = b² − 4ac
b = −4, a = 2, c = 5
D = (−4)² − 4(2)(5) = 16 − 40 = −24
→ D < 0 means _____ real roots
b = −4, a = 2, c = 5
D = (−4)² − 4(2)(5) = 16 − 40 = −24
→ D < 0 means _____ real roots
Explanation
The discriminant D = b² − 4ac = 16 − 40 = −24.
Since D < 0, the quadratic has no real solutions — only two complex (imaginary) roots.
Memory key: D < 0 → No real roots | D = 0 → One root | D > 0 → Two roots
Since D < 0, the quadratic has no real solutions — only two complex (imaginary) roots.
Memory key: D < 0 → No real roots | D = 0 → One root | D > 0 → Two roots
02
⚡ VERTEX FORM: y = a(x−h)²+k
The vertex of y = 3(x + 4)² − 7 is:
Pattern Recognition
y = a(x − h)² + k → vertex = (h, k)
y = 3(x − (−4))² + (−7)
⚠ Watch the sign of h! x + 4 = x − (−4)
y = 3(x − (−4))² + (−7)
⚠ Watch the sign of h! x + 4 = x − (−4)
Explanation
In vertex form y = a(x − h)² + k, the vertex is (h, k).
Rewrite: y = 3(x − (−4))² + (−7), so h = −4, k = −7.
Trap: The most common mistake is reading x + 4 as h = +4. Always flip the sign inside the parentheses!
Rewrite: y = 3(x − (−4))² + (−7), so h = −4, k = −7.
Trap: The most common mistake is reading x + 4 as h = +4. Always flip the sign inside the parentheses!
Polynomials & Factoring
03
⚡ REMAINDER THEOREM: f(c) = remainder
When f(x) = x³ − 2x² + 5x − 3 is divided by (x − 2), the remainder is:
Remainder Theorem — No Long Division Needed!
Divisor = (x − c) → just compute f(c)
(x − 2) means c = 2
f(2) = (2)³ − 2(2)² + 5(2) − 3
f(2) = 8 − 8 + 10 − 3 = ?
(x − 2) means c = 2
f(2) = (2)³ − 2(2)² + 5(2) − 3
f(2) = 8 − 8 + 10 − 3 = ?
Explanation
By the Remainder Theorem, the remainder when dividing by (x − 2) equals f(2).
f(2) = 8 − 8 + 10 − 3 = 7
Bonus: since f(2) ≠ 0, (x − 2) is not a factor of f(x).
f(2) = 8 − 8 + 10 − 3 = 7
Bonus: since f(2) ≠ 0, (x − 2) is not a factor of f(x).
04
⚡ RATIONAL ROOT THEOREM: ±(factors of a₀ / factors of aₙ)
Which of the following is a possible rational zero of f(x) = 4x³ + 3x² − 7x + 2?CONCEPT
Rational Root Test Setup
Possible rational zeros = ±(factors of constant) / (factors of leading coeff)
Constant = 2 → factors: 1, 2
Leading coeff = 4 → factors: 1, 2, 4
Possible zeros: ±1, ±2, ±1/2, ±1/4, ±3/2...
Constant = 2 → factors: 1, 2
Leading coeff = 4 → factors: 1, 2, 4
Possible zeros: ±1, ±2, ±1/2, ±1/4, ±3/2...
Explanation
Possible rational zeros: ± (factors of 2) / (factors of 4)
= ±{1, 2} / {1, 2, 4} = ±1, ±2, ±1/2, ±1/4
Only ±1/4 appears in this list. 3 is not a factor of 2, so ±3 and ±3/4 are impossible. ±4/3 would need 3 as a factor of the constant.
= ±{1, 2} / {1, 2, 4} = ±1, ±2, ±1/2, ±1/4
Only ±1/4 appears in this list. 3 is not a factor of 2, so ±3 and ±3/4 are impossible. ±4/3 would need 3 as a factor of the constant.
Exponential & Logarithmic Functions
05
⚡ LOG RULE: logₐ(aˣ) = x and aˡᵒᵍₐ⁽ˣ⁾ = x
Solve: log₂(x − 3) + log₂(x + 1) = 5TRAP
Product Rule + Domain Check
log product rule: logₐM + logₐN = logₐ(MN)
→ log₂[(x−3)(x+1)] = 5
→ (x−3)(x+1) = 2⁵ = 32
→ x² − 2x − 3 = 32
→ x² − 2x − 35 = 0
⚠ Check: argument must be POSITIVE!
→ log₂[(x−3)(x+1)] = 5
→ (x−3)(x+1) = 2⁵ = 32
→ x² − 2x − 3 = 32
→ x² − 2x − 35 = 0
⚠ Check: argument must be POSITIVE!
Explanation
x² − 2x − 35 = 0 → (x − 7)(x + 5) = 0 → x = 7 or x = −5
Domain check: x − 3 > 0 requires x > 3, and x + 1 > 0 requires x > −1.
x = −5 fails both conditions → rejected!
Only x = 7 is valid. Always check domain for log equations!
Domain check: x − 3 > 0 requires x > 3, and x + 1 > 0 requires x > −1.
x = −5 fails both conditions → rejected!
Only x = 7 is valid. Always check domain for log equations!
06
⚡ CHANGE OF BASE: logₐb = log b / log a
If 3^(2x−1) = 81, then x = ?
Same-Base Strategy
81 = 3⁴ → rewrite right side as same base
3^(2x−1) = 3⁴
Same base → exponents must be equal
2x − 1 = 4 → solve for x
3^(2x−1) = 3⁴
Same base → exponents must be equal
2x − 1 = 4 → solve for x
Explanation
81 = 3⁴, so the equation becomes 3^(2x−1) = 3⁴.
Equal bases → equal exponents: 2x − 1 = 4 → 2x = 5 → x = 5/2.
Equal bases → equal exponents: 2x − 1 = 4 → 2x = 5 → x = 5/2.
Complex Numbers & Rational Expressions
07
⚡ i cycle: i¹=i i²=−1 i³=−i i⁴=1 (repeats!)
Simplify: i²⁷TRAP
Cycle of 4 Method
Divide exponent by 4, check remainder:
27 ÷ 4 = 6 remainder 3
So i²⁷ = i³ = ?
i¹=i, i²=−1, i³=−i, i⁴=1
27 ÷ 4 = 6 remainder 3
So i²⁷ = i³ = ?
i¹=i, i²=−1, i³=−i, i⁴=1
Explanation
The powers of i repeat in a cycle of 4: i, −1, −i, 1, i, −1, −i, 1, ...
27 ÷ 4 = 6 remainder 3 → same as i³ = −i.
Memory: remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.
27 ÷ 4 = 6 remainder 3 → same as i³ = −i.
Memory: remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.
Sequences, Series & Probability
08
⚡ GEOMETRIC SUM: Sₙ = a₁(1−rⁿ)/(1−r) if r≠1
Find the sum of the geometric series: 3 + 6 + 12 + 24 + ... + 384
Step-by-Step Setup
First term a₁ = 3, common ratio r = 2
Find n: 3 · 2^(n−1) = 384 → 2^(n−1) = 128 = 2⁷ → n = 8
Sₙ = a₁(1 − rⁿ) / (1 − r)
S₈ = 3(1 − 2⁸) / (1 − 2)
Find n: 3 · 2^(n−1) = 384 → 2^(n−1) = 128 = 2⁷ → n = 8
Sₙ = a₁(1 − rⁿ) / (1 − r)
S₈ = 3(1 − 2⁸) / (1 − 2)
Explanation
n = 8 terms (verify: 3, 6, 12, 24, 48, 96, 192, 384 ✓)
S₈ = 3(1 − 2⁸)/(1 − 2) = 3(1 − 256)/(−1) = 3(−255)/(−1) = 765
S₈ = 3(1 − 2⁸)/(1 − 2) = 3(1 − 256)/(−1) = 3(−255)/(−1) = 765
09
⚡ BINOMIAL THEOREM: C(n,k)·aⁿ⁻ᵏ·bᵏ
What is the coefficient of x²y³ in the expansion of (x + y)⁵?CONCEPT
Binomial Term Formula
Term with x^(n−k) · y^k is: C(n, k) · x^(n−k) · y^k
We want x²y³: so n−k=2 and k=3 → n=5 ✓
Coefficient = C(5, 3) = 5! / (3! · 2!) = ?
We want x²y³: so n−k=2 and k=3 → n=5 ✓
Coefficient = C(5, 3) = 5! / (3! · 2!) = ?
Explanation
For x²y³ in (x+y)⁵: use k = 3 (the y-exponent).
C(5,3) = 5!/(3!·2!) = 120/12 = 10
Pascal's Row 5: 1, 5, 10, 10, 5, 1 — the 4th term (index 3) is 10.
C(5,3) = 5!/(3!·2!) = 120/12 = 10
Pascal's Row 5: 1, 5, 10, 10, 5, 1 — the 4th term (index 3) is 10.
10
⚡ INVERSE FUNCTION: swap x&y, then solve for y
Find the inverse of f(x) = (2x + 3) / (x − 1)TRAP
Inverse Procedure
Step 1: Replace f(x) with y
Step 2: Swap x and y → x = (2y + 3)/(y − 1)
Step 3: Solve for y:
x(y − 1) = 2y + 3
xy − x = 2y + 3
xy − 2y = x + 3 → y(x − 2) = x + 3
y = (x + 3)/(x − 2)
Step 2: Swap x and y → x = (2y + 3)/(y − 1)
Step 3: Solve for y:
x(y − 1) = 2y + 3
xy − x = 2y + 3
xy − 2y = x + 3 → y(x − 2) = x + 3
y = (x + 3)/(x − 2)
Explanation
Swap x and y: x = (2y+3)/(y−1)
Multiply: xy − x = 2y + 3
Group y terms: xy − 2y = x + 3 → y(x−2) = x+3
f⁻¹(x) = (x+3)/(x−2), domain: x ≠ 2.
Multiply: xy − x = 2y + 3
Group y terms: xy − 2y = x + 3 → y(x−2) = x+3
f⁻¹(x) = (x+3)/(x−2), domain: x ≠ 2.
Part II
Geometry
Triangles — Similarity & Congruence
11
⚡ TRIANGLE MIDSEGMENT = HALF the parallel side
In △ABC, DE is a midsegment parallel to BC. If BC = 18, find DE.CONCEPT
Midsegment Theorem
Midsegment: connects midpoints of two sides
Midsegment = (1/2) × parallel base
DE = (1/2) × BC = (1/2) × 18 = ?
Also: DE ∥ BC (parallel)
Midsegment = (1/2) × parallel base
DE = (1/2) × BC = (1/2) × 18 = ?
Also: DE ∥ BC (parallel)
Explanation
By the Triangle Midsegment Theorem, a midsegment is always half the length of the parallel side.
DE = (1/2) × BC = (1/2) × 18 = 9
DE = (1/2) × BC = (1/2) × 18 = 9
12
⚡ SIMILAR △: ratios of sides EQUAL, angles EQUAL
If △ABC ~ △DEF with AB = 8, DE = 12, and the area of △ABC = 32, find the area of △DEF.TRAP
Scale Factor vs Area Ratio
Side ratio (scale factor) k = DE/AB = 12/8 = 3/2
Area ratio = k² (NOT k!)
Area(DEF)/Area(ABC) = (3/2)² = 9/4
Area(DEF) = 32 × (9/4) = ?
Area ratio = k² (NOT k!)
Area(DEF)/Area(ABC) = (3/2)² = 9/4
Area(DEF) = 32 × (9/4) = ?
Explanation
Key trap: When figures are similar, areas scale by the square of the linear scale factor.
Scale factor = 12/8 = 3/2, so area ratio = (3/2)² = 9/4.
Area(DEF) = 32 × 9/4 = 72
Scale factor = 12/8 = 3/2, so area ratio = (3/2)² = 9/4.
Area(DEF) = 32 × 9/4 = 72
Circles — Arcs, Chords & Angles
13
⚡ INSCRIBED ANGLE = (1/2) × intercepted arc
An inscribed angle intercepts an arc of 110°. Find the inscribed angle measure.CONCEPT
Inscribed Angle Theorem
Inscribed angle = (1/2) × intercepted arc
Central angle = intercepted arc (same!)
So inscribed angle = (1/2) × central angle
Angle = (1/2) × 110° = ?
Central angle = intercepted arc (same!)
So inscribed angle = (1/2) × central angle
Angle = (1/2) × 110° = ?
Explanation
The Inscribed Angle Theorem: inscribed angle = (1/2) × intercepted arc.
(1/2) × 110° = 55°
Contrast: a central angle equals its intercepted arc exactly.
(1/2) × 110° = 55°
Contrast: a central angle equals its intercepted arc exactly.
14
⚡ TWO CHORDS CROSS: AE·EC = BE·ED
Two chords AB and CD intersect inside a circle at point E. AE = 4, EB = 9, CE = 6. Find ED.TRAP
Intersecting Chords Theorem
When two chords cross inside a circle:
(segment 1a)(segment 1b) = (segment 2a)(segment 2b)
AE · EB = CE · ED
4 · 9 = 6 · ED
36 = 6 · ED → ED = ?
(segment 1a)(segment 1b) = (segment 2a)(segment 2b)
AE · EB = CE · ED
4 · 9 = 6 · ED
36 = 6 · ED → ED = ?
Explanation
Intersecting Chords: AE × EB = CE × ED
4 × 9 = 6 × ED → 36 = 6 × ED → ED = 6
4 × 9 = 6 × ED → 36 = 6 × ED → ED = 6
Coordinate Geometry & Transformations
15
⚡ PERPENDICULAR LINES: slopes multiply to −1
A line passes through (2, 5) and is perpendicular to y = (3/4)x − 2. Find its equation.CONCEPT
Perpendicular Slope Rule
Given slope m = 3/4
Perpendicular slope = −1/m = −4/3
Use point-slope: y − y₁ = m(x − x₁)
y − 5 = −4/3(x − 2)
y = −(4/3)x + 8/3 + 5 = −(4/3)x + 23/3
Perpendicular slope = −1/m = −4/3
Use point-slope: y − y₁ = m(x − x₁)
y − 5 = −4/3(x − 2)
y = −(4/3)x + 8/3 + 5 = −(4/3)x + 23/3
Explanation
Original slope = 3/4. Perpendicular slope = −4/3 (negative reciprocal).
Point-slope through (2, 5): y − 5 = −(4/3)(x − 2)
y = −(4/3)x + 8/3 + 15/3 = −(4/3)x + 23/3
Point-slope through (2, 5): y − 5 = −(4/3)(x − 2)
y = −(4/3)x + 8/3 + 15/3 = −(4/3)x + 23/3
Area, Volume & 3D Geometry
16
⚡ PYRAMID VOLUME = (1/3) × Base Area × Height
A square pyramid has base side 6 and height 8. Find its volume.CONCEPT
Volume Formula
V = (1/3) × B × h
Base area B = 6² = 36
V = (1/3) × 36 × 8 = ?
⚠ Don't forget the 1/3! (Most missed step)
Base area B = 6² = 36
V = (1/3) × 36 × 8 = ?
⚠ Don't forget the 1/3! (Most missed step)
Explanation
V = (1/3) × base area × height = (1/3) × 36 × 8 = 96
Most common error: forgetting the 1/3 factor. A pyramid is exactly 1/3 the volume of a prism with the same base and height.
Most common error: forgetting the 1/3 factor. A pyramid is exactly 1/3 the volume of a prism with the same base and height.
17
⚡ SECTOR AREA = (θ/360) × πr²
A circle has radius 10. Find the area of a sector with central angle 72°. Use π ≈ 3.14.CONCEPT
Sector = Slice of Pie
Think of it as a fraction of the full circle:
Fraction = 72/360 = 1/5
Full circle area = π × 10² = 314
Sector area = (1/5) × 314 = ?
Fraction = 72/360 = 1/5
Full circle area = π × 10² = 314
Sector area = (1/5) × 314 = ?
Explanation
Sector Area = (72/360) × π × 10² = (1/5) × 3.14 × 100 = (1/5) × 314 = 62.8
Angle Relationships & Parallel Lines
18
⚡ EXTERIOR ANGLE = sum of 2 NON-ADJACENT interior angles
In a triangle, two interior angles are 47° and 63°. What is the exterior angle adjacent to the third interior angle?TRAP
Exterior Angle Theorem vs. Supplementary Approach
Method 1 (Direct): Exterior ∠ = sum of 2 remote interior ∠s
= 47° + 63° = 110°
Method 2 (Verify): Third interior ∠ = 180°−47°−63° = 70°
Exterior ∠ = 180° − 70° = 110° ✓
= 47° + 63° = 110°
Method 2 (Verify): Third interior ∠ = 180°−47°−63° = 70°
Exterior ∠ = 180° − 70° = 110° ✓
Explanation
Exterior Angle Theorem: an exterior angle of a triangle = sum of the two non-adjacent (remote) interior angles.
Exterior angle = 47° + 63° = 110°
Exterior angle = 47° + 63° = 110°
19
⚡ PYTHAGOREAN TRIPLE shortcut: 3-4-5, 5-12-13, 8-15-17
In right △PQR with hypotenuse PR = 26 and leg PQ = 10, find leg QR.CONCEPT
Pythagorean Theorem
a² + b² = c² (c = hypotenuse)
10² + QR² = 26²
100 + QR² = 676
QR² = 576 → QR = √576 = ?
Hint: is this a scaled Pythagorean triple?
10² + QR² = 26²
100 + QR² = 676
QR² = 576 → QR = √576 = ?
Hint: is this a scaled Pythagorean triple?
Explanation
QR² = 26² − 10² = 676 − 100 = 576 → QR = 24
This is the 5-12-13 triple scaled by 2: (10, 24, 26) = 2 × (5, 12, 13). Recognizing triples saves time!
This is the 5-12-13 triple scaled by 2: (10, 24, 26) = 2 × (5, 12, 13). Recognizing triples saves time!
20
⚡ TANGENT-SECANT from exterior: (whole × external) = (whole × external)
From external point P, a secant through the circle has external segment 4 and total length 9. A tangent from P has length t. Find t.TRAP
Power of a Point
Tangent-Secant rule (from external point):
tangent² = external segment × whole secant
t² = 4 × 9
t² = 36 → t = ?
⚠ Use WHOLE secant length (not internal chord only)
tangent² = external segment × whole secant
t² = 4 × 9
t² = 36 → t = ?
⚠ Use WHOLE secant length (not internal chord only)
Explanation
Power of a Point (tangent-secant): t² = external × whole secant
t² = 4 × 9 = 36 → t = 6
Trap: Using only the internal chord (5) instead of the full secant (9) gives the wrong answer.
t² = 4 × 9 = 36 → t = 6
Trap: Using only the internal chord (5) instead of the full secant (9) gives the wrong answer.