Part 1
Algebra 2
KEY: discriminant = b²−4ac → tells you HOW MANY real roots
How many real solutions does the equation \(2x^2 - 4x + 3 = 0\) have?
✦ Quick Check
For \(ax^2+bx+c=0\), compute \(\Delta = b^2-4ac\).
• \(\Delta > 0\) → 2 real roots • \(\Delta = 0\) → 1 real root • \(\Delta < 0\) → no real roots
• \(\Delta > 0\) → 2 real roots • \(\Delta = 0\) → 1 real root • \(\Delta < 0\) → no real roots
📖 Explanation
Compute \(\Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8\).
Since \(\Delta < 0\), there are no real solutions. The correct answer is A.
Tip: Negative discriminant → no real roots (but two complex roots exist).
Since \(\Delta < 0\), there are no real solutions. The correct answer is A.
Tip: Negative discriminant → no real roots (but two complex roots exist).
KEY: vertex form y = a(x−h)²+k → vertex is (h, k) NOT (−h, k)!
The parabola \(y = -3(x+2)^2 + 7\) has its vertex at:
✦ Watch Out!
If the form is \((x \mathbf{+} 2)^2\), think of it as \((x - (\mathbf{-2}))^2\).
So \(h = -2\), not \(+2\). This is the most common mistake!
So \(h = -2\), not \(+2\). This is the most common mistake!
📖 Explanation
In \(y = a(x-h)^2+k\), the vertex is \((h,k)\).
Rewrite: \(y = -3(x-(-2))^2+7\), so \(h=-2\), \(k=7\).
Vertex = \(\mathbf{(-2,\ 7)}\). Answer: B.
Rewrite: \(y = -3(x-(-2))^2+7\), so \(h=-2\), \(k=7\).
Vertex = \(\mathbf{(-2,\ 7)}\). Answer: B.
KEY: remainder theorem → plug x=c into f(x) directly
When \(f(x) = x^3 - 4x^2 + 5x - 2\) is divided by \((x - 1)\), what is the remainder?
✦ Remainder Theorem
Dividing \(f(x)\) by \((x-c)\) → remainder = \(f(c)\).
No long division needed! Just substitute \(x = c\).
No long division needed! Just substitute \(x = c\).
📖 Explanation
\(f(1) = 1^3 - 4(1)^2 + 5(1) - 2 = 1 - 4 + 5 - 2 = 0\).
Remainder = \(\mathbf{0}\). This also means \((x-1)\) is a factor of \(f(x)\). Answer: A.
Remainder = \(\mathbf{0}\). This also means \((x-1)\) is a factor of \(f(x)\). Answer: A.
KEY: log rule → log(A·B) = logA + logB | log(Aⁿ) = n·logA
Simplify: \(\log_2 8 + \log_2 4\)
✦ Two Methods
Method 1 (Product Rule): \(\log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32\)
Method 2 (Direct): \(2^3=8\) so \(\log_2 8=3\); \(2^2=4\) so \(\log_2 4=2\). Then \(3+2=5\).
Method 2 (Direct): \(2^3=8\) so \(\log_2 8=3\); \(2^2=4\) so \(\log_2 4=2\). Then \(3+2=5\).
📖 Explanation
\(\log_2 8 = 3\) (since \(2^3=8\)) and \(\log_2 4 = 2\) (since \(2^2=4\)).
\(3 + 2 = \mathbf{5}\). Answer: C.
\(3 + 2 = \mathbf{5}\). Answer: C.
KEY: same base → set exponents equal!
Solve for \(x\): \(3^{2x-1} = 27\)
✦ Strategy
Express both sides with the same base: \(27 = 3^3\).
Then \(3^{2x-1} = 3^3 \Rightarrow 2x-1 = 3\).
Then \(3^{2x-1} = 3^3 \Rightarrow 2x-1 = 3\).
📖 Explanation
\(27 = 3^3\), so \(3^{2x-1} = 3^3 \Rightarrow 2x-1=3 \Rightarrow 2x=4 \Rightarrow x=2\).
Check: \(3^{2(2)-1} = 3^3 = 27\) ✓. Answer: B.
Check: \(3^{2(2)-1} = 3^3 = 27\) ✓. Answer: B.
KEY: factor first, then cancel COMMON factors (not individual terms!)
Simplify: \(\dfrac{x^2 - 9}{x^2 - x - 6}\) where \(x \neq 3, -2\)
✦ Common Mistake
You CANNOT cancel the \(x^2\) from numerator and denominator as if they were separate terms.
You MUST factor both completely first: \(x^2-9 = (x-3)(x+3)\), \(x^2-x-6 = (x-3)(x+2)\).
You MUST factor both completely first: \(x^2-9 = (x-3)(x+3)\), \(x^2-x-6 = (x-3)(x+2)\).
📖 Explanation
Factor: \(\dfrac{(x-3)(x+3)}{(x-3)(x+2)}\).
Cancel \((x-3)\): result is \(\dfrac{x+3}{x+2}\). Answer: B.
Cancel \((x-3)\): result is \(\dfrac{x+3}{x+2}\). Answer: B.
KEY: substitution or elimination → always CHECK answer in BOTH equations
Solve the system: \(\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}\)
✦ Elimination Tip
Add both equations to eliminate \(y\): \((2x+y) + (x-y) = 7+2 \Rightarrow 3x = 9\).
📖 Explanation
Adding: \(3x = 9 \Rightarrow x=3\). Then \(y = x-2 = 1\).
Check: \(2(3)+1=7\) ✓ and \(3-1=2\) ✓. Answer: B.
Check: \(2(3)+1=7\) ✓ and \(3-1=2\) ✓. Answer: B.
KEY: i² = −1 always! Replace i² with −1 when simplifying.
Compute \((3 + 2i)(1 - i)\).
✦ FOIL it
\((3)(1) + (3)(-i) + (2i)(1) + (2i)(-i)\)
\(= 3 - 3i + 2i - 2i^2\) then use \(i^2 = -1\).
\(= 3 - 3i + 2i - 2i^2\) then use \(i^2 = -1\).
📖 Explanation
FOIL: \(3 - 3i + 2i - 2i^2 = 3 - i - 2(-1) = 3 - i + 2 = 5 - i\). Answer: C.
KEY: composite function → f(g(x)) means plug g(x) INTO f(x)
If \(f(x) = x^2 + 1\) and \(g(x) = 2x - 3\), find \(f(g(2))\).
✦ Order Matters!
Step 1: Find \(g(2)\) first. Step 2: Plug that result into \(f\).
Many students accidentally do \(g(f(2))\) — always work from the inside out!
Many students accidentally do \(g(f(2))\) — always work from the inside out!
📖 Explanation
\(g(2) = 2(2)-3 = 1\).
\(f(g(2)) = f(1) = 1^2+1 = \mathbf{2}\). Answer: A.
\(f(g(2)) = f(1) = 1^2+1 = \mathbf{2}\). Answer: A.
KEY: nth term → aₙ = a₁ + (n−1)d | "d" = common difference
In an arithmetic sequence, the 3rd term is 11 and the 7th term is 23. What is the 1st term?
✦ Find d first
From term 3 to term 7 is 4 steps, and the value increases by \(23-11=12\).
So \(d = 12 \div 4 = 3\). Then work backwards: \(a_1 = a_3 - 2d\).
So \(d = 12 \div 4 = 3\). Then work backwards: \(a_1 = a_3 - 2d\).
📖 Explanation
\(d = \frac{23-11}{7-3} = \frac{12}{4} = 3\).
\(a_1 = a_3 - 2d = 11 - 6 = \mathbf{5}\).
Check: 5, 8, 11, 14, 17, 20, 23 ✓. Answer: C.
\(a_1 = a_3 - 2d = 11 - 6 = \mathbf{5}\).
Check: 5, 8, 11, 14, 17, 20, 23 ✓. Answer: C.
Part 2
Geometry
KEY: triangle angle sum = 180° always!
In a triangle, two angles measure \(47°\) and \(68°\). What is the third angle?
✦ Rule
Sum of all interior angles of any triangle = \(180°\).
Third angle = \(180° - 47° - 68°\).
Third angle = \(180° - 47° - 68°\).
📖 Explanation
\(180° - 47° - 68° = \mathbf{65°}\). Answer: C.
KEY: a² + b² = c² where c is ALWAYS the HYPOTENUSE (longest side)
A right triangle has legs of length \(5\) and \(12\). What is the length of the hypotenuse?
✦ Classic Triple
\(5, 12, 13\) is a Pythagorean triple — memorize it!
Other common ones: \(3,4,5\) and \(8,15,17\).
Other common ones: \(3,4,5\) and \(8,15,17\).
📖 Explanation
\(c = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = \mathbf{13}\). Answer: B.
KEY: Area = πr² | Circumference = 2πr | r = radius, NOT diameter!
A circle has a diameter of 10. What is its area?
✦ Don't Forget!
Diameter = 10 → radius = 5. Many students forget to halve the diameter before using the formula.
📖 Explanation
\(r = 10 \div 2 = 5\). Area \(= \pi r^2 = \pi(5)^2 = \mathbf{25\pi}\). Answer: B.
Note: \(100\pi\) is the trap answer when you forget to halve the diameter.
Note: \(100\pi\) is the trap answer when you forget to halve the diameter.
KEY: alternate interior angles = EQUAL | co-interior (same-side) angles = 180°
Two parallel lines are cut by a transversal. One co-interior angle (same-side interior) measures \(112°\). What is the other co-interior angle?
✦ Angle Pairs
Alternate interior: equal | Corresponding: equal | Co-interior (consecutive): supplementary = 180°
"Co-interior = supplementary" is the most confused rule.
"Co-interior = supplementary" is the most confused rule.
📖 Explanation
Co-interior angles are supplementary: \(180° - 112° = \mathbf{68°}\). Answer: B.
KEY: similar triangles → corresponding sides are PROPORTIONAL (set up a ratio!)
Two similar triangles have corresponding sides in the ratio \(3:5\). The area of the smaller triangle is \(27\ \text{cm}^2\). What is the area of the larger triangle?
✦ Area Scale Factor
If the side ratio is \(k\), then the area ratio is \(k^2\).
Side ratio = \(3:5\) → area ratio = \(3^2 : 5^2 = 9 : 25\).
Side ratio = \(3:5\) → area ratio = \(3^2 : 5^2 = 9 : 25\).
📖 Explanation
Area ratio = \(9:25\). So larger area = \(27 \times \dfrac{25}{9} = 3 \times 25 = \mathbf{75}\ \text{cm}^2\). Answer: C.
KEY: midpoint = average of coordinates → M = ((x₁+x₂)/2, (y₁+y₂)/2)
Point \(M(3, -1)\) is the midpoint of segment \(\overline{AB}\). If \(A = (7, 4)\), find point \(B\).
✦ Work Backwards
\(3 = \frac{7+x_B}{2} \Rightarrow x_B = 6-7 = -1\)
\(-1 = \frac{4+y_B}{2} \Rightarrow y_B = -2-4 = -6\)
\(-1 = \frac{4+y_B}{2} \Rightarrow y_B = -2-4 = -6\)
📖 Explanation
\(x_B = 2(3)-7 = -1\), \(y_B = 2(-1)-4 = -6\). So \(B = (-1,\ -6)\). Answer: A.
KEY: Volume of cone = (1/3)πr²h | cylinder = πr²h → cone is 1/3 of cylinder!
A cone has a radius of \(6\) cm and a height of \(9\) cm. What is its volume?
(Leave answer in terms of \(\pi\))
(Leave answer in terms of \(\pi\))
✦ Memory Hack
Three cones fit inside one cylinder of the same base and height.
So \(V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}}\).
So \(V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}}\).
📖 Explanation
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(36)(9) = \frac{1}{3}(324\pi) = \mathbf{108\pi}\ \text{cm}^3\). Answer: C.
Note: \(324\pi\) is what you get without the \(\frac{1}{3}\) — the most common error.
Note: \(324\pi\) is what you get without the \(\frac{1}{3}\) — the most common error.
KEY: inscribed angle = HALF of its intercepted arc (central angle)
An inscribed angle in a circle intercepts an arc of \(140°\). What is the measure of the inscribed angle?
✦ Inscribed vs Central
Central angle = arc measure (same) ← vertex at center
Inscribed angle = arc ÷ 2 ← vertex on the circle
Inscribed angle = arc ÷ 2 ← vertex on the circle
📖 Explanation
Inscribed angle = \(\frac{1}{2} \times \text{arc} = \frac{1}{2} \times 140° = \mathbf{70°}\). Answer: B.
KEY: reflection over y-axis → (x, y) becomes (−x, y) | over x-axis → (x, −y)
Point \(P(4, -3)\) is reflected over the y-axis. What are the coordinates of its image \(P'\)?
✦ Quick Reference
Reflect over y-axis: flip the x-sign only → \((x,y) \to (-x, y)\)
Reflect over x-axis: flip the y-sign only → \((x,y) \to (x, -y)\)
Reflect over x-axis: flip the y-sign only → \((x,y) \to (x, -y)\)
📖 Explanation
Reflecting over the y-axis negates the x-coordinate: \((4,-3) \to (-4,-3)\). Answer: C.
KEY: SOH-CAH-TOA → sin=Opp/Hyp, cos=Adj/Hyp, tan=Opp/Adj
In a right triangle, the angle \(\theta = 30°\), and the hypotenuse is \(10\). What is the length of the side opposite to \(\theta\)?
✦ 30-60-90 Triangle
Sides are in ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse)
\(\sin 30° = \frac{1}{2}\) → opposite = hypotenuse × \(\frac{1}{2}\)
\(\sin 30° = \frac{1}{2}\) → opposite = hypotenuse × \(\frac{1}{2}\)
📖 Explanation
\(\sin 30° = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}\).
Opposite \(= 10 \times \frac{1}{2} = \mathbf{5}\). Answer: B.
Note: \(5\sqrt{3}\) is the adjacent side (cos 30° × 10) — watch out!
Opposite \(= 10 \times \frac{1}{2} = \mathbf{5}\). Answer: B.
Note: \(5\sqrt{3}\) is the adjacent side (cos 30° × 10) — watch out!