Self-Study Guide Β· Math Mastery

Algebra 2
& Geometry

20 essential problems. Choose your answer and check instantly. Common mistakes explained.

10 Algebra 2 10 Geometry Memory Points Instant Feedback
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Algebra 2 10 QUESTIONS
Q 01
Quadratic Formula
DISCRIMINANT & NATURE OF ROOTS
⚑ MEMORY: bΒ²βˆ’4ac β†’ (+) two real, (0) one real, (βˆ’) no real
Example
For \(x^2 - 4x + 4 = 0\): discriminant \(= (-4)^2 - 4(1)(4) = 0\) β†’ exactly one real root (double root): \(x = 2\)
What is the discriminant of \(2x^2 - 3x + 5 = 0\), and how many real solutions does it have?
πŸ“˜ Explanation
\(\Delta = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31\). Since \(\Delta < 0\), there are no real solutions (two complex/imaginary roots). Trap: students often forget to apply the sign of \(c\) correctly. Here \(c = +5\), so \(4ac = 40\).
Q 02
Complex Numbers
POWERS OF i
⚑ MEMORY: i cycle β†’ i, βˆ’1, βˆ’i, 1 (period 4). Divide exponent by 4, use remainder.
Example
\(i^{13}\): \(13 \div 4 = 3\) remainder \(1\) β†’ \(i^1 = i\)
Simplify \(i^{47} + i^{50}\).
πŸ“˜ Explanation
\(47 \div 4 = 11\) R\(3\) β†’ \(i^{47} = i^3 = -i\).
\(50 \div 4 = 12\) R\(2\) β†’ \(i^{50} = i^2 = -1\).
So \(i^{47} + i^{50} = -i + (-1) = -i - 1\). Answer: C. Common trap: confusing remainder 2 with \(-1\) vs \(+1\).
Q 03
Polynomial Functions
END BEHAVIOR
⚑ MEMORY: Leading term decides all. Even degree β†’ same ends. Odd β†’ opposite ends. Negative flips.
Example
\(f(x) = -x^4 + 3x\): even degree, negative leading β†’ both ends go to \(-\infty\)
Describe the end behavior of \(f(x) = -3x^5 + 7x^2 - 2\).
πŸ“˜ Explanation
Leading term: \(-3x^5\). Odd degree + negative leading coefficient β†’ as \(x \to +\infty\), \(f \to -\infty\); as \(x \to -\infty\), \(f \to +\infty\). Answer: B. Without the negative, odd-degree goes up-right and down-left. The negative flips both.
Q 04
Logarithms
LOG PROPERTIES β€” COMMON MISTAKE
⚑ MEMORY: log(a/b) = log a βˆ’ log b. NOT log a / log b ← WRONG!
Example
\(\log_2 8 - \log_2 2 = \log_2(8/2) = \log_2 4 = 2\)
Which expression is equal to \(\log_3(27x^4)\)?
πŸ“˜ Explanation
\(\log_3(27x^4) = \log_3 27 + \log_3 x^4 = 3 + 4\log_3 x\).
Because \(3^3 = 27\) so \(\log_3 27 = 3\). Power rule: \(\log x^4 = 4\log x\). Answer: A. Trap: B confuses \(\log_3 27 = 3\) with \(= 27\).
Q 05
Rational Exponents
FRACTIONAL EXPONENT ↔ RADICAL
⚑ MEMORY: x^(m/n) = (ⁿ√x)^m. Bottom of fraction = root index. Top = power.
Example
\(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\)
Evaluate \((-27)^{2/3}\).
πŸ“˜ Explanation
\((-27)^{2/3} = (\sqrt[3]{-27})^2 = (-3)^2 = 9\). Answer: A. Trap: Many write \(-9\) because they see the negative sign and keep it. But squaring always gives a positive result. Note: cube roots of negatives ARE defined (unlike even roots of negatives).
Q 06
Sequences & Series
GEOMETRIC SERIES SUM
⚑ MEMORY: S = aΒ·(1βˆ’rⁿ)/(1βˆ’r). Infinite geometric (|r|<1): S = a/(1βˆ’r)
Example
Sum of \(2 + 6 + 18 + 54\): \(a=2, r=3, n=4\) β†’ \(S = 2\cdot\frac{1-3^4}{1-3} = 2\cdot\frac{-80}{-2} = 80\)
Find the sum of the infinite geometric series: \(12 + 4 + \dfrac{4}{3} + \cdots\)
πŸ“˜ Explanation
Common ratio: \(r = 4/12 = 1/3\). Since \(|r| = 1/3 < 1\), the series converges.
\(S_\infty = \frac{a}{1-r} = \frac{12}{1 - 1/3} = \frac{12}{2/3} = 12 \times \frac{3}{2} = 18\). Answer: B. Trap: A (16) comes from adding only first 3 terms. Always use the formula.
Q 07
Radical Equations
EXTRANEOUS SOLUTIONS
⚑ MEMORY: After solving radical equation β†’ ALWAYS CHECK back in original. Squaring can create fake solutions!
Example
\(\sqrt{x+3} = x-3\) β†’ square both: \(x+3 = x^2-6x+9\) β†’ \(x^2-7x+6=0\) β†’ \(x=6\) or \(x=1\). Check: \(x=1\): \(\sqrt{4}=2\) but \(1-3=-2\). βœ— Extraneous! Only \(x=6\).
Solve \(\sqrt{3x + 4} = x - 2\). Which value(s) are valid solutions?
πŸ“˜ Explanation
Square both sides: \(3x+4 = (x-2)^2 = x^2-4x+4\) β†’ \(x^2-7x = 0\) β†’ \(x(x-7) = 0\) β†’ \(x=0\) or \(x=7\).
Check \(x=0\): \(\sqrt{4} = 2\), but \(0-2 = -2\). βœ— Extraneous.
Check \(x=7\): \(\sqrt{25} = 5\), and \(7-2 = 5\). βœ“
Answer: B (\(x=7\), not 5 β€” note the options are intentionally set with \(x=5\) as a trap).
Q 08
Conic Sections
CIRCLE EQUATION β€” STANDARD FORM
⚑ MEMORY: (xβˆ’h)Β²+(yβˆ’k)Β²=rΒ². Center=(h,k), radius=r. Complete the square to find!
Example
\(x^2+y^2-4x+6y=3\) β†’ \((x-2)^2+(y+3)^2=16\). Center: \((2,-3)\), radius: \(4\)
What is the center and radius of the circle \(x^2 + y^2 - 6x + 8y - 11 = 0\)?
πŸ“˜ Explanation
Complete the square:
\((x^2-6x+9)+(y^2+8y+16) = 11+9+16 = 36\)
\((x-3)^2+(y+4)^2 = 36\)
Center: \((3,-4)\), radius: \(\sqrt{36}=6\). Answer: A. Trap C: \(r^2=36\) but \(r=6\), not 36!
Q 09
Exponential Equations
SOLVE BY MATCHING BASES
⚑ MEMORY: Same base β†’ set exponents equal. Different base β†’ take log of both sides.
Example
\(4^x = 8\) β†’ \(2^{2x} = 2^3\) β†’ \(2x = 3\) β†’ \(x = 1.5\)
Solve for \(x\): \(9^{x+1} = 27^{x-1}\)
πŸ“˜ Explanation
Convert to base 3: \(9 = 3^2,\; 27 = 3^3\)
\(3^{2(x+1)} = 3^{3(x-1)}\)
\(2x+2 = 3x-3\)
\(x = 5\). Answer: A. Common mistake: failing to distribute the exponent β€” e.g., writing \(3^{2x+1}\) instead of \(3^{2x+2}\).
Q 10
Systems of Equations
NON-LINEAR SYSTEM β€” SUBSTITUTION
⚑ MEMORY: Line + Parabola can have 0, 1, or 2 intersections. Substitute linear into quadratic.
Example
Line \(y=x+2\) and parabola \(y=x^2\): sub β†’ \(x+2=x^2\) β†’ \(x^2-x-2=0\) β†’ \(x=2,-1\)
How many intersection points do \(y = x^2 - 4\) and \(y = 2x - 1\) have?
πŸ“˜ Explanation
Set equal: \(x^2-4 = 2x-1\) β†’ \(x^2-2x-3=0\) β†’ \((x-3)(x+1)=0\) β†’ \(x=3\) or \(x=-1\).
Two distinct solutions β†’ 2 intersection points. Answer: C. Check discriminant: \(\Delta = 4+12=16>0\) β†’ two real roots. βœ“
Geometry 10 QUESTIONS
G 01
Triangle Congruence
SSS / SAS / ASA / AAS / HL
⚑ MEMORY: SSA is NOT a congruence rule (it's the "ambiguous case"). AAA only proves similarity.
Example
Two triangles share a side (common side), have one pair of equal angles adjacent to it, and another pair of equal angles β†’ ASA congruence.
In β–³ABC and β–³DEF, AB = DE, BC = EF, and ∠B = ∠E. Which congruence postulate applies?
πŸ“˜ Explanation
We have: Side AB = DE, Angle ∠B = ∠E (the angle between the two sides), Side BC = EF. This is Side-Angle-Side (SAS). Answer: B. The angle must be included between the two sides for SAS.
G 02
Parallel Lines & Transversal
ANGLE RELATIONSHIPS
⚑ MEMORY: Co-interior (same-side interior) = supplementary (add to 180°). Alternate = equal. Corresponding = equal.
Example
Lines \(\ell \parallel m\), transversal cuts them. Alternate interior angles: \(70Β°\) β†’ other side also \(70Β°\). Co-interior: \(70Β° + ? = 180Β°\) β†’ \(? = 110Β°\).
Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle is \((3x + 15)Β°\) and the other is \((2x + 25)Β°\). Find \(x\).
πŸ“˜ Explanation
Co-interior angles are supplementary: \((3x+15)+(2x+25)=180\)
\(5x + 40 = 180\) β†’ \(5x = 140\) β†’ \(x = 28\). Answer: A. Trap: If students mistake them for alternate interior angles (setting them equal), they get \(x = 10\) β€” that's option D.
G 03
Pythagorean Theorem
RIGHT TRIANGLE β€” FIND MISSING SIDE
⚑ MEMORY: a²+b²=c² where c is ALWAYS the hypotenuse (longest side, opposite right angle).
Example
Legs: 5 and 12. Hypotenuse: \(\sqrt{25+144}=\sqrt{169}=13\). Memorize: (3,4,5), (5,12,13), (8,15,17) triples!
In a right triangle, one leg is 7 and the hypotenuse is 25. What is the other leg?
πŸ“˜ Explanation
\(a^2 + 7^2 = 25^2\) β†’ \(a^2 = 625 - 49 = 576\) β†’ \(a = 24\). Answer: A. This is the (7, 24, 25) Pythagorean triple. Common error: adding instead of subtracting β€” \(7^2 + 25^2\) gives B.
G 04
Circle Theorems
INSCRIBED ANGLE THEOREM
⚑ MEMORY: Inscribed angle = Β½ Γ— (intercepted arc). Central angle = arc. Inscribed = half of central!
Example
Arc = 140Β° β†’ Central angle = 140Β° β†’ Inscribed angle intercepting same arc = 70Β°
An inscribed angle in a circle intercepts an arc of 110Β°. What is the measure of the inscribed angle?
πŸ“˜ Explanation
Inscribed Angle Theorem: inscribed angle = \(\frac{1}{2}\) Γ— intercepted arc = \(\frac{110Β°}{2} = 55Β°\). Answer: C. Trap A: confusing inscribed angle with central angle (central = arc). Trap D: subtracting from 90Β° β€” wrong approach.
G 05
Triangle Similarity
AA / SAS~ / SSS~ + SCALE FACTOR
⚑ MEMORY: Similar β†’ angles equal, sides proportional. Area ratio = (scale factor)Β². Volume ratio = (scale factor)Β³.
Example
Scale factor 1:3 β†’ Area ratio = 1:9. If small area is 5, large area is 45.
β–³ABC ~ β–³DEF with a scale factor of 1:4. If the area of β–³ABC is 9 cmΒ², what is the area of β–³DEF?
πŸ“˜ Explanation
Area ratio = (scale factor)Β² = \(4^2 = 16\). Area of β–³DEF = \(9 \times 16 = 144\text{ cm}^2\). Answer: B. Trap A: multiplying area by 4 (not 4Β²). Always square the linear scale factor for area!
G 06
Volume of Solids
SPHERE / CONE / CYLINDER
⚑ MEMORY: Sphere=4/3Ο€rΒ³, Cone=1/3Ο€rΒ²h, Cylinder=Ο€rΒ²h. Cone = 1/3 of cylinder with same base & height!
Example
Cylinder r=3, h=10: \(V = \pi(9)(10) = 90\pi\). Cone same dimensions: \(V = 30\pi\).
A cone has radius 6 cm and height 9 cm. What is its volume? (Leave in terms of \(\pi\).)
πŸ“˜ Explanation
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6^2)(9) = \frac{1}{3}\pi(36)(9) = \frac{324\pi}{3} = 108\pi\). Answer: B. Trap A: forgetting to multiply by \(\frac{1}{3}\). Trap C: using \(r=6\) as diameter instead.
G 07
Coordinate Geometry
MIDPOINT & DISTANCE FORMULA
⚑ MEMORY: Midpoint = average the coordinates. Distance = Pythagorean theorem on the coordinate plane.
Example
Midpoint of \((1,3)\) and \((5,7)\): \(\left(\frac{1+5}{2}, \frac{3+7}{2}\right) = (3, 5)\)
Point M is the midpoint of segment AB. If A = (2, βˆ’3) and M = (5, 1), find the coordinates of B.
πŸ“˜ Explanation
Midpoint formula: \(\frac{x_A+x_B}{2}=5\) β†’ \(x_B = 10-2 = 8\). \(\frac{-3+y_B}{2}=1\) β†’ \(y_B = 2+3 = 5\). So \(B=(8,5)\). Answer: A. Trap B: averaging A and M instead of solving for B. Trap C: just adding the differences once instead of doubling.
G 08
Special Right Triangles
30-60-90 and 45-45-90
⚑ MEMORY: 45-45-90: sides = x, x, x√2. 30-60-90: sides = x, x√3, 2x. (short leg is x)
Example
30-60-90 with hypotenuse 10: short leg = 5, long leg = \(5\sqrt{3}\)
In a 45-45-90 triangle, the hypotenuse is \(8\sqrt{2}\). What is the length of each leg?
πŸ“˜ Explanation
45-45-90 rule: if leg = \(x\), hypotenuse = \(x\sqrt{2}\). So \(x\sqrt{2} = 8\sqrt{2}\) β†’ \(x = 8\). Each leg = \(8\). Answer: A. Trap B: dividing by \(\sqrt{2}\) incorrectly. Trap D: dividing the entire expression by 2.
G 09
Polygon Interior Angles
SUM OF INTERIOR ANGLES
⚑ MEMORY: Sum of interior angles = (nβˆ’2)Γ—180Β°. Each angle of regular polygon = (nβˆ’2)Γ—180Β°/n
Example
Hexagon (n=6): \((6-2)\times180 = 720Β°\). Each angle of regular hexagon: \(720/6 = 120Β°\)
What is the measure of each interior angle of a regular nonagon (9 sides)?
πŸ“˜ Explanation
Sum = \((9-2)\times180 = 7\times180 = 1260Β°\). Each angle = \(1260/9 = 140Β°\). Answer: B. Trap A: that's a hexagon. Trap D: that's an octagon. Trap C: that's an 18-gon. Know your regular polygons!
G 10
Transformations
ROTATIONS β€” 90Β°, 180Β°, 270Β°
⚑ MEMORY: Rotate 90Β° CCW: (x,y)β†’(βˆ’y,x). 180Β°: (x,y)β†’(βˆ’x,βˆ’y). 270Β° CCW = 90Β° CW: (x,y)β†’(y,βˆ’x)
Example
Point (3, 2) rotated 90Β° CCW about origin: β†’ \((-2, 3)\)
Point P(4, βˆ’3) is rotated 270Β° counterclockwise about the origin. What are the new coordinates?
πŸ“˜ Explanation
270Β° CCW = 90Β° CW: rule is \((x,y) \to (y, -x)\).
P(4, βˆ’3) β†’ \((-3, -4)\). Answer: A. Check: 90Β° CCW gives \((3,4)\), 180Β° gives \((-4,3)\), 270Β° CCW gives \((-3,-4)\). Memorize the three rotation rules!
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