Algebra 1 Solving Quadratic Equations Test Solutions and Scoring (PDF)

This algebra 1 quadratic equations test answers for Algebra 1 covers solving quadratic equations (Common Core A-REI.B.4a, A-REI.B.4b, A-SSE.B.3a, A-CED.A.1), with a free PDF to print.

Algebra 1 · Common Core State Standards: A-REI.B.4a, A-REI.B.4b, A-SSE.B.3a, A-CED.A.1
Algebra 1 quadratic equations test answers: worked solutions
Question 1 – Square roots [2 pts]
a) \(x^{2} = 64\), so \(x = \pm 8\).
b) \(x^{2} = 25\), so \(x = \pm 5\).
\(x = \pm 8\); \(x = \pm 5\)
Scoring: 1 point for each part (both solutions).
Question 2 – Factoring [2 pts]
\((x + 6)(x – 4) = 0\).
\(x = -6\) or \(x = 4\)
Scoring: 1 point for the factors; 1 point for both solutions.
Question 3 – Factoring with a leading coefficient [2 pts]
\((3x – 1)(x – 3) = 0\).
\(x = \frac{1}{3}\) or \(x = 3\).
\(x = \frac{1}{3}\) or \(x = 3\)
Scoring: 1 point for the factors; 1 point for both solutions.
Question 4 – Completing the square [2 pts]
Move the constant: \(x^{2} + 8x = -5\). Add 16: \((x + 4)^{2} = 11\).
\(x = -4 \pm \sqrt{11}\)
Scoring: 1 point for completing the square correctly; 1 point for the solutions.
Question 5 – Quadratic formula [2 pts]
\(a = 3, b = -2, c = -4\). Discriminant: \(4 + 48 = 52\), and \(\sqrt{52} = 2\sqrt{13}\).
\(x = \frac{2 \pm 2\sqrt{13}}{6} = \frac{1 \pm \sqrt{13}}{3}\).
\(x = \frac{1 \pm \sqrt{13}}{3}\)
Scoring: 1 point for the discriminant; 1 point for the simplified solutions.
Question 6 – Discriminant [2 pts]
\(x^{2} + 4x + 7 = 0\): \(16 – 28 = -12\), no real solutions.
\(4x^{2} – 12x + 9 = 0\): \(144 – 144 = 0\), one real solution.
-12, none; 0, one
Scoring: 1 point for each equation.
Question 7 – Rectangle area [3 pts]
a) \(w(w + 5) = 84\), so \(w^{2} + 5w – 84 = 0\).
b) \((w + 12)(w – 7) = 0\), so \(w = -12\) or \(w = 7\).
c) A width cannot be negative: width 7 m, length 12 m.
Width 7 m, length 12 m
Scoring: 1 point per part.
Question 8 – Ball thrown from a platform [3 pts]
a) \(-16t^{2} + 40t + 24 = 0\) divided by \(-8\) gives \(2t^{2} – 5t – 3 = 0\).
b) \((2t + 1)(t – 3) = 0\), so \(t = -\frac{1}{2}\) or \(t = 3\).
c) Time cannot be negative, so the ball lands at 3 seconds.
3 seconds
Scoring: 1 point per part.
Question 9 – Find the error [2 pts]
The student forgot the negative square root.
\(x – 2 = \pm 5\), so \(x = 7\) or \(x = -3\).
\(x = 7\) or \(x = -3\)
Scoring: 1 point for the explanation; 1 point for both solutions.
Total score
20 points
Key facts behind this algebra 1 quadratic equations test answers
- Standard form: ax² + bx + c = 0 with a ≠ 0.
- Zero product property: if A·B = 0, then A = 0 or B = 0.
- Square roots: x² = k gives x = ±√k.
- Quadratic formula: x = (-b ± √(b² – 4ac)) / 2a.
- The discriminant b² – 4ac: positive for two real solutions, zero for one, negative for none.
How to use this algebra 1 quadratic equations test answers
Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.
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