Algebra 2 & Geometry — Study Guide

Algebra 2

10 problems · Quadratics · Functions · Logarithms · Systems

Quadratic Formula · Discriminant

DISCRIMINANT: b²−4ac → + two roots · 0 one root · − no real roots

Worked Example

For x² − 5x + 6 = 0, the discriminant is 25 − 24 = 1 > 0, so there are two real roots: x = 3 and x = 2.

Your Turn

How many real solutions does the equation 2x² − 4x + 5 = 0 have?

Discriminant = b² - 4ac = (-4)² - 4(2)(5)

💡 Explanation Calculate: b² − 4ac = 16 − 40 = −24. Since the discriminant is negative, the equation has no real solutions. The parabola never touches the x-axis. Tip: many students forget to multiply 4·a·c, computing only b² by mistake.

Vertex Form · Completing the Square

VERTEX FORM: y = a(x−h)² + k → vertex is (h, k), watch the SIGN of h!

Worked Example

y = (x − 3)² + 2 → vertex is (3, 2). The minus sign inside means h = +3, not −3.

The parabola y = −2(x + 1)² − 4 has its vertex at which point?

💡 Explanation The form is y = a(x − h)² + k. Rewrite: y = −2(x − (−1))² + (−4). So h = −1 and k = −4. The vertex is (−1, −4). The most common mistake: reading (x + 1) as h = +1 instead of h = −1.

Logarithm Laws · Expanding Logs

LOG RULES: log(AB)=logA+logB · log(A/B)=logA−logB · log(Aⁿ)=n·logA

Worked Example

Expand: log₂(8x³) = log₂8 + log₂(x³) = 3 + 3log₂x

Which expression is equal to log₃(27x²)?

💡 Explanation log₃(27x²) = log₃27 + log₃(x²) = log₃(3³) + 2log₃x = 3 + 2log₃x. Students often forget to apply the power rule to x², writing log₃x instead of 2log₃x.

Exponential Equations · Solving with Same Base

SAME BASE: if aˣ = aʸ then x = y · convert both sides to same base first!

Worked Example

Solve 4ˣ = 8: rewrite as (2²)ˣ = 2³, so 2x = 3, giving x = 3/2.

Solve for x: 9ˣ⁺¹ = 27

💡 Explanation Rewrite: 9ˣ⁺¹ = (3²)ˣ⁺¹ = 3²⁽ˣ⁺¹⁾ and 27 = 3³. So 2(x+1) = 3 → 2x + 2 = 3 → 2x = 1 → x = 1/2. Trap: students pick x=1 by guessing 9² = 81 ≠ 27.

Rational Expressions · Restrictions on Domain

UNDEFINED: denominator = 0 → set bottom = 0, solve, those values are EXCLUDED

What value(s) of x must be excluded from the domain of

(x + 3) / (x² - x - 6)

💡 Explanation Factor the denominator: x² − x − 6 = (x − 3)(x + 2). Set each factor = 0: x = 3 and x = −2. These are the excluded values. Note: even if (x+3) cancels with the numerator, x = 3 is still excluded from the original domain.

Function Composition · f(g(x))

COMPOSITION: f(g(x)) means plug g(x) INTO f · work INSIDE OUT

Worked Example

If f(x) = x² + 1 and g(x) = 2x, then f(g(x)) = f(2x) = (2x)² + 1 = 4x² + 1.

Given f(x) = 3x − 1 and g(x) = x², what is f(g(2))?

💡 Explanation Step 1: g(2) = 2² = 4. Step 2: f(g(2)) = f(4) = 3(4) − 1 = 11. Common error: computing f(2) first, then squaring, which gives g(f(2)) instead — order matters!

Arithmetic Sequences · nth Term Formula

ARITHMETIC: aₙ = a₁ + (n−1)d · find d FIRST (difference of consecutive terms)

The first term of an arithmetic sequence is 7, and the common difference is −3. What is the 15th term?

aₙ = a₁ + (n - 1)d

💡 Explanation a₁₅ = 7 + (15 − 1)(−3) = 7 + 14(−3) = 7 − 42 = −35. A very common mistake: using n = 15 instead of (n−1) = 14, which gives 7 + 15(−3) = −38 (wrong!). Always subtract 1.

Systems of Equations · Substitution / Elimination

ELIMINATION: multiply to make coefficients equal, then ADD or SUBTRACT equations

Solve the system:

3x + 2y = 12 5x - 2y = 4

💡 Explanation Add the two equations: (3x + 2y) + (5x − 2y) = 12 + 4 → 8x = 16 → x = 2. Substitute back: 3(2) + 2y = 12 → 2y = 6 → y = 3. Answer: (2, 3). Always verify in BOTH equations!

Polynomial Long Division · Remainder Theorem

REMAINDER THEOREM: remainder of f(x)÷(x−a) equals f(a) — just PLUG IN a!

Worked Example

The remainder when f(x) = x³ − 2x + 1 is divided by (x − 2) is f(2) = 8 − 4 + 1 = 5.

What is the remainder when f(x) = 2x³ + x² − 5x + 3 is divided by (x − 1)?

💡 Explanation By the Remainder Theorem, substitute x = 1: f(1) = 2(1)³ + (1)² − 5(1) + 3 = 2 + 1 − 5 + 3 = 1. No long division needed! Students who do the full division often make arithmetic errors. Use the shortcut.

A10

Complex Numbers · i² = −1

IMAGINARY: i = √(−1), i² = −1, i³ = −i, i⁴ = 1 → cycle repeats every 4!

Simplify: (3 + 2i)(1 − 4i)

Use FOIL: (a+bi)(c+di) = ac + adi + bci + bdi²

💡 Explanation FOIL: 3(1) + 3(−4i) + 2i(1) + 2i(−4i) = 3 − 12i + 2i − 8i². Since i² = −1: = 3 − 10i − 8(−1) = 3 − 10i + 8 = 11 − 10i. Critical step: replace i² with −1, not +1. This sign flip trips up most students.

Geometry

10 problems · Triangles · Circles · Area & Volume · Proofs

Triangle Similarity · AA Postulate

AA SIMILARITY: 2 angles equal → triangles similar → corresponding sides PROPORTIONAL

Worked Example

If △ABC ~ △DEF with ratio 1:3, and AB = 5, then DE = 15. Scale all sides by the same ratio.

Two triangles are similar. The sides of the smaller triangle are 4, 6, and 8. The longest side of the larger triangle is 20. What is the shortest side of the larger triangle?

💡 Explanation The scale factor = 20 ÷ 8 = 2.5. The shortest side = 4 × 2.5 = 10. Always match longest-to-longest and shortest-to-shortest. Picking the wrong corresponding sides is the #1 mistake here.

Pythagorean Theorem · Special Right Triangles

45-45-90: sides = x, x, x√2 · 30-60-90: sides = x, x√3, 2x (hyp = 2×short!)

In a 30-60-90 triangle, the hypotenuse is 14. What is the length of the side opposite the 30° angle?

💡 Explanation In a 30-60-90 triangle: hypotenuse = 2 × (short side). So short side = 14 ÷ 2 = 7. The side opposite 60° would be 7√3. Students commonly confuse which ratio belongs to which angle.

Circle Theorems · Inscribed Angle

INSCRIBED ANGLE = half the intercepted arc · CENTRAL ANGLE = full arc · ratio is 1:2

Worked Example

If arc AB = 80°, then inscribed angle = 40°. Central angle = 80°.

An inscribed angle in a circle intercepts an arc of 110°. What is the measure of the inscribed angle?

💡 Explanation Inscribed Angle Theorem: inscribed angle = ½ × intercepted arc. So ½ × 110° = 55°. Students often confuse this with the central angle (which equals the arc directly). Remember: inscribed = HALF.

Area of Composite Figures

COMPOSITE: BREAK into simple shapes → ADD or SUBTRACT their areas

A rectangle is 10 cm wide and 8 cm tall. A semicircle with diameter 10 cm sits on top of it. What is the total area? (Use π ≈ 3.14)

Area = rectangle + semicircle = (l\timesw) + (πr²/2)

💡 Explanation Rectangle = 10 × 8 = 80 cm². Semicircle: diameter = 10, so radius = 5. Area = π(5²)/2 = 3.14 × 25/2 = 39.25 cm². Total = 80 + 39.25 = 119.25 cm². Trap: using the diameter (10) as the radius.

Volume of 3D Solids · Cone vs Cylinder

CONE = ⅓ × cylinder of same base & height · V = (1/3)πr²h

A cone has a radius of 6 cm and a height of 9 cm. What is its volume? (Use π ≈ 3.14)

💡 Explanation V = (1/3)πr²h = (1/3) × 3.14 × 36 × 9 = (1/3) × 1017.36 = 339.12 cm³. Choice A is the cylinder volume (forgetting the 1/3). This is the most common error — always multiply by 1/3 for cones!

Parallel Lines · Transversal Angle Relationships

PARALLEL LINES: alternate interior = EQUAL · co-interior (same side) = 180° · corresponding = EQUAL

Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle measures 73°. What is the measure of its co-interior partner?

💡 Explanation Co-interior (consecutive interior) angles are supplementary: they add up to 180°. So 180° − 73° = 107°. Students often confuse this with alternate interior angles (which are equal). Co-interior = supplementary (180°).

Coordinate Geometry · Midpoint & Distance

MIDPOINT: average the x's, average the y's → M = ((x₁+x₂)/2, (y₁+y₂)/2)

Point A is at (−2, 5) and Point B is at (6, −3). What is the midpoint of segment AB?

💡 Explanation x: (−2 + 6)/2 = 4/2 = 2. y: (5 + (−3))/2 = 2/2 = 1. Midpoint = (2, 1). Common error: subtracting instead of adding the coordinates, or forgetting to divide by 2.

Polygon Interior Angles · Sum Formula

SUM of interior angles = (n−2) × 180° · each angle in REGULAR polygon = sum ÷ n

What is the measure of each interior angle of a regular octagon?

Sum = (n - 2) \times 180°, then divide by n

💡 Explanation Octagon has n = 8 sides. Sum = (8 − 2) × 180° = 6 × 180° = 1080°. Each angle = 1080° ÷ 8 = 135°. Note: 120° is a regular hexagon, 144° is a regular decagon. Don't memorize — use the formula each time.

Trigonometry · SOH-CAH-TOA in Right Triangles

SOH: sin = opp/hyp · CAH: cos = adj/hyp · TOA: tan = opp/adj

Worked Example

In a right triangle, angle θ = 30°, hypotenuse = 10. The opposite side = 10 × sin(30°) = 10 × 0.5 = 5.

In a right triangle, one acute angle is 40°. The side adjacent to this angle is 12. What is the hypotenuse? (cos 40° ≈ 0.766)

💡 Explanation cos(40°) = adjacent/hypotenuse → 0.766 = 12/h → h = 12/0.766 ≈ 15.7. Trap: multiplying instead of dividing (12 × 0.766 ≈ 9.2). When you know the adjacent and want the hypotenuse, you DIVIDE by cos.

G10

Transformations · Reflection · Rotation Rules

ROTATION 90° CCW: (x,y) → (−y, x) · 180°: (x,y) → (−x,−y) · 270° CCW: (x,y) → (y,−x)

Point P is at (3, −5). It is rotated 180° about the origin. What are the new coordinates?

💡 Explanation A 180° rotation: (x, y) → (−x, −y). So (3, −5) → (−3, −(−5)) = (−3, 5). Many students negate only one coordinate. For 180°, both x and y change sign.