PART 01
Algebra 2 — Word Problems
🧠 KEY: DISCRIMINANT · VERTEX · INVERSE · ASYMPTOTE · COMPLETE-THE-SQUARE
📘 Quick Tip
For a quadratic \(ax^2+bx+c=0\), the discriminant is \(\Delta = b^2-4ac\).• \(\Delta > 0\) → 2 real roots | \(\Delta = 0\) → 1 real root | \(\Delta < 0\) → no real roots
A ball is thrown upward from a 4-foot platform. Its height in feet after \(t\) seconds is modeled by
\[h(t) = -16t^2 + 24t + 4\]
How many times does the ball reach a height of 16 feet?
Step-by-step
Set \(h(t)=16\): \(-16t^2+24t+4=16 \Rightarrow -16t^2+24t-12=0 \Rightarrow 4t^2-6t+3=0\).
Discriminant: \(\Delta = 36 - 48 = -12 < 0\). No real roots → the ball never reaches 16 ft.
⚠ Common trap: Students forget to subtract 16 from both sides first, then misread the discriminant sign.
Discriminant: \(\Delta = 36 - 48 = -12 < 0\). No real roots → the ball never reaches 16 ft.
⚠ Common trap: Students forget to subtract 16 from both sides first, then misread the discriminant sign.
📘 Quick Tip
Vertex formula: \(x = -\dfrac{b}{2a}\), then substitute back for \(y\).Memory word: VERTEX = MAX/MIN point
A farmer has 200 meters of fence to enclose a rectangular field against a straight river (no fence needed along the river). What is the maximum area the farmer can enclose?
Step-by-step
Let width \(= w\). Two widths + one length = 200 → \(L = 200 - 2w\).
Area: \(A = w(200-2w) = 200w - 2w^2\).
Vertex at \(w = \frac{200}{4} = 50\), so \(L = 100\).
Maximum area = \(50 \times 100 = \mathbf{5{,}000}\) m².
Area: \(A = w(200-2w) = 200w - 2w^2\).
Vertex at \(w = \frac{200}{4} = 50\), so \(L = 100\).
Maximum area = \(50 \times 100 = \mathbf{5{,}000}\) m².
📘 Quick Tip
\(\log_b x = y \Leftrightarrow b^y = x\). Always check: argument of log must be > 0.
A population of bacteria doubles every 3 hours. Starting with 500 bacteria, after how many hours will the population first exceed 8,000?
(Use \(\log_2 16 = 4\))
Step-by-step
Model: \(P(t) = 500 \cdot 2^{t/3}\). Solve \(500 \cdot 2^{t/3} > 8000\).
\(2^{t/3} > 16 = 2^4 \Rightarrow t/3 > 4 \Rightarrow t > 12\).
First whole hour exceeding is \(t = \mathbf{12}\) hours (exactly 8,000 at 12, exceeds after).
⚠ Trap: confusing "first exceeds" with "equals."
\(2^{t/3} > 16 = 2^4 \Rightarrow t/3 > 4 \Rightarrow t > 12\).
First whole hour exceeding is \(t = \mathbf{12}\) hours (exactly 8,000 at 12, exceeds after).
⚠ Trap: confusing "first exceeds" with "equals."
📘 Quick Tip
When solving radical equations: isolate the radical → square both sides → CHECK all answers (extraneous solutions appear after squaring).
A square garden has an area of \((x^2 + 10x + 25)\) square feet. A gardener adds a 3-foot-wide border around it. What is the total area of the garden including the border when \(x = 2\)?
Step-by-step
\(x^2+10x+25 = (x+5)^2\). Side of garden = \(x+5\). At \(x=2\): side = 7 ft.
With 3-ft border on each side: new side = \(7 + 3 + 3 = 13\) ft.
Total area = \(13^2 = \mathbf{169}\) ft².
With 3-ft border on each side: new side = \(7 + 3 + 3 = 13\) ft.
Total area = \(13^2 = \mathbf{169}\) ft².
📘 Quick Tip
Rational function \(\dfrac{k}{x-a}+b\): vertical asymptote at \(x=a\), horizontal asymptote at \(y=b\).Memory: VERTICAL = denominator zero · HORIZONTAL = end behavior
A company's average cost per unit (in dollars) is modeled by
\[C(x) = \frac{1200}{x} + 15\]
where \(x\) is the number of units produced. As production increases without bound, the average cost approaches what value?
Step-by-step
As \(x \to \infty\): \(\dfrac{1200}{x} \to 0\), so \(C(x) \to 15\).
The horizontal asymptote is \(y = \mathbf{15}\). Average cost approaches $15.
The horizontal asymptote is \(y = \mathbf{15}\). Average cost approaches $15.
📘 Quick Tip
To find \(f^{-1}(x)\): swap \(x\) and \(y\), then solve for \(y\).Memory: INVERSE = SWAP + SOLVE
A Celsius-to-Fahrenheit converter uses \(F = \frac{9}{5}C + 32\). A scientist needs to convert 77°F back to Celsius. Using the inverse function, what is the result?
Step-by-step
Inverse: \(C = \dfrac{5}{9}(F-32)\).
\(C = \dfrac{5}{9}(77-32) = \dfrac{5}{9}(45) = \dfrac{225}{9} = \mathbf{25}\)°C.
⚠ Trap: Many students subtract 32 after multiplying. Remember: subtract first.
\(C = \dfrac{5}{9}(77-32) = \dfrac{5}{9}(45) = \dfrac{225}{9} = \mathbf{25}\)°C.
⚠ Trap: Many students subtract 32 after multiplying. Remember: subtract first.
📘 Quick Tip
Completing the square: \(x^2+bx = \left(x+\dfrac{b}{2}\right)^2 - \dfrac{b^2}{4}\)Memory: HALF · SQUARE · ADD-SUBTRACT
A drone's flight path follows \(h = -t^2 + 6t - 5\) (height in meters, time in seconds). What is the maximum height the drone reaches?
Step-by-step
Complete the square: \(h = -(t^2-6t)-5 = -[(t-3)^2-9]-5 = -(t-3)^2+4\).
Vertex: \((3, 4)\) → maximum height = 4 meters at \(t=3\) sec.
Vertex: \((3, 4)\) → maximum height = 4 meters at \(t=3\) sec.
📘 Quick Tip
Change of base: \(\log_b a = \dfrac{\ln a}{\ln b}\).Memory: LOG RULES: Product → ADD · Quotient → SUBTRACT · Power → MULTIPLY
An earthquake measures 6.0 on the Richter scale. A second earthquake is 100 times more intense. What is the Richter magnitude of the second earthquake?
(Richter scale: \(M = \log_{10} I\) where \(I\) is intensity)
Step-by-step
New intensity = \(100 \times I_1\).
New magnitude = \(\log_{10}(100 \cdot I_1) = \log_{10}100 + \log_{10}I_1 = 2 + 6.0 = \mathbf{8.0}\).
⚠ Trap: "100 times" = +2 on log scale, not ×2.
New magnitude = \(\log_{10}(100 \cdot I_1) = \log_{10}100 + \log_{10}I_1 = 2 + 6.0 = \mathbf{8.0}\).
⚠ Trap: "100 times" = +2 on log scale, not ×2.
📘 Quick Tip
Quadratic formula: \(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)Memory: NEGATIVE B, PLUS-MINUS ROOT, OVER 2A
Two numbers have a sum of 10 and a product of 21. Which of the following correctly represents both numbers?
Step-by-step
Let the numbers be \(r\) and \(s\). Then \(r+s=10\), \(rs=21\).
They are roots of \(t^2-10t+21=0 = (t-3)(t-7)\).
Solutions: \(t = 3\) and \(t = 7\). ✓ \(3+7=10\), \(3 \times 7=21\).
They are roots of \(t^2-10t+21=0 = (t-3)(t-7)\).
Solutions: \(t = 3\) and \(t = 7\). ✓ \(3+7=10\), \(3 \times 7=21\).
📘 Quick Tip
Compound interest: \(A = P\!\left(1+\dfrac{r}{n}\right)^{nt}\)Continuous: \(A = Pe^{rt}\) · Memory: PERT
You invest $1,000 at an annual interest rate of 6%, compounded annually. After 2 years, which expression gives the correct balance?
Step-by-step
Annual compounding: \(n=1\), so formula = \(P(1+r)^t = 1000(1.06)^2 = \mathbf{\$1{,}123.60}\).
Choice B is simple interest ×2. Choice C is also simple interest. Choice D is continuous compounding.
Choice B is simple interest ×2. Choice C is also simple interest. Choice D is continuous compounding.
PART 02
Geometry — Core Problems
📐 KEY: SIMILAR · PARALLEL · CONGRUENT · PYTHAGOREAN · ARC-SECTOR · INSCRIBED-ANGLE
📘 Quick Tip
Similar triangles: corresponding sides are proportional.Memory: SIMILAR = SAME SHAPE, DIFFERENT SIZE
Triangle ABC is similar to Triangle DEF. In △ABC, \(AB = 6\), \(BC = 9\), and \(AC = 12\). If \(DE = 4\), what is the length of \(EF\)?
Step-by-step
Scale factor: \(\dfrac{DE}{AB} = \dfrac{4}{6} = \dfrac{2}{3}\).
\(EF = BC \times \dfrac{2}{3} = 9 \times \dfrac{2}{3} = \mathbf{6}\).
⚠ Trap: must match corresponding sides AB↔DE, BC↔EF.
\(EF = BC \times \dfrac{2}{3} = 9 \times \dfrac{2}{3} = \mathbf{6}\).
⚠ Trap: must match corresponding sides AB↔DE, BC↔EF.
📘 Quick Tip
Inscribed angle = half the intercepted arc.Central angle = the intercepted arc.
Memory: INSCRIBED = HALF
In circle O, an inscribed angle \(\angle ABC\) intercepts an arc of 110°. What is the measure of \(\angle ABC\)?
Step-by-step
Inscribed Angle Theorem: inscribed angle = \(\frac{1}{2}\) × intercepted arc.
\(\angle ABC = \frac{1}{2} \times 110° = \mathbf{55°}\).
\(\angle ABC = \frac{1}{2} \times 110° = \mathbf{55°}\).
📘 Quick Tip
Arc length: \(s = r\theta\) (radians) or \(s = \dfrac{\theta}{360°} \times 2\pi r\)Sector area: \(A = \dfrac{\theta}{360°} \times \pi r^2\)
Memory: ARC = FRACTION of CIRCUMFERENCE
A pizza slice is a sector of a circle with radius 10 inches and central angle 60°. What is the area of the slice? (Use \(\pi \approx 3.14\))
Step-by-step
\(A = \dfrac{60}{360} \times \pi \times 10^2 = \dfrac{1}{6} \times 3.14 \times 100 = \dfrac{314}{6} \approx \mathbf{52.3}\) in².
📘 Quick Tip
Parallel lines cut by a transversal:• Alternate interior angles: EQUAL
• Co-interior (same-side) angles: SUPPLEMENTARY (sum = 180°)
Memory: Z-angles EQUAL · C-angles 180°
Two parallel lines are cut by a transversal. One co-interior angle measures \((3x + 20)°\) and the other measures \((x + 40)°\). Find the value of \(x\).
Step-by-step
Co-interior angles are supplementary: \((3x+20)+(x+40)=180\).
\(4x + 60 = 180 \Rightarrow 4x = 120 \Rightarrow x = \mathbf{30}\).
Check: \(110°+70°=180°\) ✓
\(4x + 60 = 180 \Rightarrow 4x = 120 \Rightarrow x = \mathbf{30}\).
Check: \(110°+70°=180°\) ✓
📘 Quick Tip
Pythagorean theorem: \(a^2 + b^2 = c^2\) (c = hypotenuse).Common triples: 3-4-5 · 5-12-13 · 8-15-17
A ladder 13 feet long leans against a vertical wall. The base of the ladder is 5 feet from the wall. How high up the wall does the ladder reach?
Step-by-step
\(h^2 + 5^2 = 13^2 \Rightarrow h^2 = 169 - 25 = 144 \Rightarrow h = \mathbf{12}\) feet.
Recognize the 5-12-13 Pythagorean triple!
Recognize the 5-12-13 Pythagorean triple!
📘 Quick Tip
Cylinder total surface area: \(SA = 2\pi r^2 + 2\pi rh\)Memory: TWO CIRCLES + RECTANGLE (unrolled)
A cylindrical water tank has a radius of 3 m and a height of 10 m. What is the total surface area of the tank, including both circular ends? (Leave answer in terms of \(\pi\))
Step-by-step
\(SA = 2\pi(3)^2 + 2\pi(3)(10) = 18\pi + 60\pi = \mathbf{78\pi}\) m².
⚠ Trap: "lateral area only" = \(60\pi\). Always re-read whether ends are included.
⚠ Trap: "lateral area only" = \(60\pi\). Always re-read whether ends are included.
📘 Quick Tip
Triangle Midsegment Theorem: A segment connecting midpoints of two sides is(1) parallel to the third side and (2) half its length.
Memory: MIDSEGMENT = HALF the BASE
In △PQR, M is the midpoint of PQ and N is the midpoint of PR. If MN = 14 cm, what is the length of QR?
Step-by-step
By the Midsegment Theorem: \(MN = \frac{1}{2} QR\).
So \(QR = 2 \times MN = 2 \times 14 = \mathbf{28}\) cm.
⚠ Common error: students divide instead of multiply.
So \(QR = 2 \times MN = 2 \times 14 = \mathbf{28}\) cm.
⚠ Common error: students divide instead of multiply.
📘 Quick Tip
Exterior Angle Theorem: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.Memory: EXTERIOR = SUM OF REMOTE INTERIORS
In △ABC, the exterior angle at C measures 115°. If one of the non-adjacent interior angles (at A) is 65°, what is the measure of the interior angle at B?
Step-by-step
Exterior Angle Theorem: \(\angle A + \angle B = 115°\).
\(65° + \angle B = 115° \Rightarrow \angle B = \mathbf{50°}\).
\(65° + \angle B = 115° \Rightarrow \angle B = \mathbf{50°}\).
📘 Quick Tip
Cone volume: \(V = \frac{1}{3}\pi r^2 h\)Sphere volume: \(V = \frac{4}{3}\pi r^3\)
Memory: CONE = ⅓ CYLINDER · SPHERE = 4/3 πr³
An ice cream cone has radius 3 cm and height 12 cm. A perfectly spherical scoop of ice cream with radius 3 cm sits on top. What is the total volume? (Leave in terms of \(\pi\))
Step-by-step
Cone: \(V_c = \frac{1}{3}\pi(3)^2(12) = \frac{1}{3}\pi \cdot 9 \cdot 12 = 36\pi\)
Sphere: \(V_s = \frac{4}{3}\pi(3)^3 = \frac{4}{3}\pi \cdot 27 = 36\pi\)
Total = \(36\pi + 36\pi = \mathbf{72\pi}\) cm³.
Sphere: \(V_s = \frac{4}{3}\pi(3)^3 = \frac{4}{3}\pi \cdot 27 = 36\pi\)
Total = \(36\pi + 36\pi = \mathbf{72\pi}\) cm³.
📘 Quick Tip
Tangent-radius: always perpendicular (90°).Two tangents from an external point: both are equal in length.
Memory: TANGENT ⊥ RADIUS · TWO TANGENTS = EQUAL
From external point P, two tangent segments PA and PB are drawn to circle O. If PA = \(3x - 1\) and PB = \(x + 7\), what is the length of each tangent?
Step-by-step
Two tangents from an external point are equal: \(PA = PB\).
\(3x-1 = x+7 \Rightarrow 2x = 8 \Rightarrow x = 4\).
Length = \(3(4)-1 = \mathbf{11}\). Check: \(4+7=11\) ✓
\(3x-1 = x+7 \Rightarrow 2x = 8 \Rightarrow x = 4\).
Length = \(3(4)-1 = \mathbf{11}\). Check: \(4+7=11\) ✓