Algebra 1 — Semester B Practice

Question 1 of 10

TEKS 6A-6C Medium Calc Word

Use the quadratic formula for x² − 4x + 3 = 0

A x = 2, 4

B x = 0, 4

C x = −1, −3

D x = 1, 3

Explanation

📌 a=1, b=−4, c=3. x = (4 ± √(16−12))/2 = (4±2)/2 → x=3 or x=1

Question 2 of 10

TEKS 1A-1G Medium Calc Word

A ball is thrown upward. Its height is h = −16t² + 48t. When does it hit the ground?

A 2 sec

B 1 sec

C 3 sec

D 4 sec

Explanation

📌 h = 0: −16t² + 48t = 0 → t(−16t + 48) = 0 → t = 0 or t = 3 seconds

Question 3 of 10

TEKS 6A-6C Hard Calc Word

Complete the square: x² + 6x + ___ = (x + ___)²

A 6, 3

B 36, 6

C 3, 9

D 9, 3

Explanation

📌 Half of 6 = 3. Square it = 9.
x² + 6x + 9 = (x + 3)²

Question 4 of 10

TEKS 6A-6C Hard Calc Word

Use quadratic formula: 2x²−5x−3=0

A x=5 or x=−3

B x=−3 or x=1/2

C x=2 or x=−3

D x=3 or x=−1/2

Explanation

📌 x=(5±√(25+24))/4=(5±7)/4. x=3 or x=−1/2

Question 5 of 10

TEKS 7A-7C Easy Calc Word

The vertex of y = (x − 3)² + 2 is:

A (3, −2)

B (3, 2)

C (−3, 2)

D (2, 3)

Explanation

📌 Vertex form: y = a(x−h)² + k. Vertex = (h, k) = (3, 2).

Question 6 of 10

TEKS 6A-6C Medium Calc Word Diagram

The parabola in the graph below has a vertex at which point?

A (1, -2)

B (0, 0)

C (2, 1)

D (-1, 2)

Explanation

The vertex is the lowest point of an upward-opening parabola, marked at approximately (1, -2).

Question 7 of 10

TEKS 6A-6C Medium Calc

Factor: x² + 5x + 6

A (x−2)(x−3)

B (x+1)(x+6)

C (x+2)(x+3)

D (x+3)(x+3)

Explanation

📌 Find two numbers that multiply to 6 and add to 5: 2 and 3.
(x + 2)(x + 3)

Question 8 of 10

TEKS 7A-7C Easy Calc Word Diagram

Which graph represents a quadratic function?

A B

B Both

C A

D Neither

Explanation

📌 Quadratic = U-shaped parabola. Graph A shows a parabola.
Graph B is a straight line → linear, not quadratic.

Question 9 of 10

TEKS 8A-8B Medium Calc Diagram

For a quadratic equation ax² + bx + c = 0, the discriminant is b² − 4ac. It tells us how many times the related parabola y = ax² + bx + c crosses the x-axis — and therefore how many real solutions the equation has.

The discriminant's sign matches the number of x-axis crossings.

Which statement about D = 0 is TRUE?

A Two real solutions — the parabola crosses the x-axis twice

B No real solutions — the parabola doesn't reach the x-axis

C The number of solutions can't be determined from D alone

D Exactly one real solution — the parabola is tangent to the x-axis

Explanation

The discriminant b² − 4ac counts how many real solutions a quadratic equation has by reflecting how many times the parabola y = ax² + bx + c crosses the x-axis.

The three cases:
• D > 0: parabola crosses the x-axis at TWO different points → 2 real solutions.
• D = 0: parabola is *tangent* to the x-axis — it touches at exactly ONE point (the vertex itself sits on the x-axis) → 1 real solution (often called a *double root*).
• D < 0: parabola sits entirely above or below the x-axis, never touching it → 0 real solutions (the roots are complex / imaginary).

Application: D = 0 is the borderline case useful for problems like "for what value of c does ax² + bx + c = 0 have exactly one solution?" — set b² − 4ac = 0 and solve.

Question 10 of 10

TEKS 1A-1G Easy Calc

A square garden has an area of 64 square feet. If the side length is x feet, the relationship is x² = 64.

The algebra gives two mathematical solutions, but only one makes physical sense as a side length. Which choice gives BOTH algebraic solutions AND correctly identifies the physical side length?

A Only x = 8 — that's the only solution to x² = 64

B x = 32, because half of 64 is the side length

C Only x = −8, because squaring a negative gives a positive

D x = ±8 algebraically; the garden's side length is +8 ft

Explanation

Taking the square root of both sides gives x = ±√64 = ±8, because both (+8)² = 64 AND (−8)² = 64. In pure algebra you must report both roots. For the physical garden, length can't be negative, so only +8 ft applies in context — but the algebraic answer set is {−8, +8}.

Common mistakes: (A) Dropping the negative root entirely. (D) Dividing 64 by 2 instead of taking a square root.

Tip: when you see x² = N, always write x = ±√N first, then decide which roots fit the real-world context.

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