Algebra 2 Solving Quadratic Equations Worksheet Solutions (PDF)
This algebra 2 quadratic equations answers for Algebra 2 covers solving quadratic equations (Common Core A-REI.B.4a, A-REI.B.4b), with a free PDF to print.
Every worksheet problem on solving quadratic equations is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where a lost plus-or-minus sign or a wrong sign in the formula can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.
Algebra 2 · Common Core State Standards: A-REI.B.4a, A-REI.B.4b
Algebra 2 quadratic equations answers: exercises 1 to 10
Exercise 1 – Square roots
Take the square root of both sides, keeping both signs.
\(x = \pm 7\)
Exercise 2 – Isolate the square
\(3x^{2} = 75\), so \(x^{2} = 25\).
\(x = \pm 5\)
Exercise 3 – Square of a binomial
\(x – 4 = \pm 6\), so \(x = 4 \pm 6\).
\(x = 10, -2\)
Exercise 4 – Irrational solutions
\(x + 2 = \pm\sqrt{11}\), so \(x = -2 \pm \sqrt{11}\).
\(x = -2 + \sqrt{11}, -2 – \sqrt{11}\)
Exercise 5 – Factor and solve
\((x – 5)(x + 2) = 0\). By the zero product property, \(x – 5 = 0\) or \(x + 2 = 0\).
\(x = 5, -2\)
Exercise 6 – Factor, a = 2
\((2x – 1)(x + 4) = 0\). Check: \(2x^{2} + 8x – x – 4\).
\(x = \frac{1}{2}, -4\)
Exercise 7 – Factoring and the graph
\((x – 3)(x – 5) = 0\).
\(x = 3, 5\), matching the x-intercepts
Exercise 8 – Complete the square, integer
\(x^{2} + 6x = 7\). Add \(9\): \(x^{2} + 6x + 9 = 16\).
\((x + 3)^{2} = 16\), so \(x + 3 = \pm 4\).
\(x = 1, -7\)
Exercise 9 – Complete the square, irrational
\(x^{2} – 4x = 3\). Add \(4\): \((x – 2)^{2} = 7\).
\(x – 2 = \pm\sqrt{7}\).
\(x = 2 \pm \sqrt{7}\)
Exercise 10 – Complete the square, odd b
\(x^{2} + 5x = -3\). Half of 5 is \(\frac{5}{2}\); square it: \(\frac{25}{4}\).
\(\left(x + \frac{5}{2}\right)^{2} = -3 + \frac{25}{4} = \frac{13}{4}\), so \(x + \frac{5}{2} = \pm\frac{\sqrt{13}}{2}\).
\(x = \frac{-5 \pm \sqrt{13}}{2}\)
Exercises 11 and up
Exercise 11 – Quadratic formula, rational
Identify \(a = 3, b = -5, c = -2\).
Discriminant: \(b^{2} – 4ac = (-5)^{2} – 4(3)(-2) = 49\).
Substitute: \(x = \frac{-(-5) \pm \sqrt{49}}{2(3)}\).
\(\sqrt{49} = 7\).
x = \(-\frac{1}{3}\) or x = \(2\)
Exercise 12 – Quadratic formula, radicals
Identify \(a = 2, b = 4, c = -3\).
Discriminant: \(b^{2} – 4ac = (4)^{2} – 4(2)(-3) = 40\).
Substitute: \(x = \frac{-(4) \pm \sqrt{40}}{2(2)}\).
\(\sqrt{40} = 2\sqrt{10}\).
x = \(\frac{-2 – \sqrt{10}}{2}\) or x = \(\frac{-2 + \sqrt{10}}{2}\)
Exercise 13 – Complex solutions
Identify \(a = 1, b = 2, c = 5\).
Discriminant: \(b^{2} – 4ac = (2)^{2} – 4(1)(5) = -16\).
Substitute: \(x = \frac{-(2) \pm \sqrt{-16}}{2(1)}\).
\(\sqrt{-16} = 4i\): the solutions are complex.
x = \(-1 + 2i\) or x = \(-1 – 2i\)
Exercise 14 – Discriminants
\(16 – 16 = 0\); \(1 – 4 = -3\); \(25 + 24 = 49\).
One real; two complex; two real
Exercise 15 – Find k
One real solution means \(D = 0\): \(k^{2} – 4(1)(9) = 0\), so \(k^{2} = 36\).
\(k = 6\) or \(k = -6\)
Exercise 16 – No real solutions
The vertex (2, 1) is above the x-axis and the parabola opens up, so it never reaches 0.
\(D = 16 – 20 = -4\). \(x = \frac{4 \pm 2i}{2}\).
\(x = 2 \pm i\)
Exercise 17 – Garden
\(w(w + 3) = 40\) gives \(w^{2} + 3w – 40 = 0\), so \((w + 8)(w – 5) = 0\).
The width cannot be negative, so \(w = 5\) and the length is \(w + 3 = 8\).
5 feet by 8 feet
Exercise 18 – Ball hits the ground
\(-16t^{2} + 48t + 4 = 0\) becomes \(4t^{2} – 12t – 1 = 0\).
\(D = 144 + 16 = 160\), \(\sqrt{160} = 4\sqrt{10}\), so \(t = \frac{12 \pm 4\sqrt{10}}{8} = \frac{3 \pm \sqrt{10}}{2}\).
The negative value is rejected. \(t = \frac{3 + \sqrt{10}}{2} \approx 3.08\).
About 3.08 seconds
Exercise 19 – Break-even
\(x^{2} – 12x + 20 = 0\), so \((x – 2)(x – 10) = 0\).
x = 2 and x = 10 (20 and 100 tickets)
Exercise 20 – Find the error
The zero product property only works when the product equals 0. Rewrite: \(x^{2} – 2x – 8 = 0\), so \((x – 4)(x + 2) = 0\).
\(x = 4, -2\)
Exercise 21 – Choose a method
\((x + 7)^{2} = 0\), so \(x = -7\). \(D = 196 – 196 = 0\).
One solution, x = -7 (repeated)
Key facts behind this algebra 2 quadratic equations answers
- Square roots: isolate the square, then take ± the root.
- Factoring: use the zero product property.
- Quadratic formula: x = (-b ± √(b² – 4ac)) / (2a).
- Discriminant > 0: two real; = 0: one real; < 0: two complex.
- Check solutions, and reject those that do not fit the context.
How to use this algebra 2 quadratic equations answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
Download the algebra 2 quadratic equations answers as a PDF with the button below and keep it with your notes.
Download the algebra 2 quadratic equations answers as a PDF with the button below and keep it with your notes.
All Algebra 2 chapters: Algebra 2 math.
More Algebra 2 math
Lessons
See all
Worksheets
See all
Tests
See all
Logarithms Test with Point Values- Zeros and Graphs of Polynomial Functions Test with Point Values
- Radical and Rational Equations Test with Point Values
- Sampling, Surveys and Experiments Test with Point Values
- The Normal Distribution Test with Point Values
- Systems of Linear and Quadratic Equations Test with Point Values
