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Study Guide11th GradeLevel 2 · Standard

Free 11th Grade The Quadratic Formula and the Discriminant

A senior study guide to solving any quadratic equation with the quadratic formula and using the discriminant to predict how many real solutions there are. Worked examples cover exact radical answers, equations with no real solutions, finding an unknown coefficient for equal roots, and proving a quadratic is always positive, with a six-question self-check and answers.

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Grade
11th Grade
Subject
Math
Topic
Quadratics
Difficulty
Level 2 · Standard
Estimated time
25 minutes
Curriculum
Texas grade levels (TEKS)
Answers
Not applicable
Format
PDF (A4) + print

Students will practice

  • solving quadratic equations with the quadratic formula, including exact radical form
  • calculating the discriminant and interpreting its sign
  • finding unknown coefficients so that an equation has equal, distinct or no real roots
  • linking the discriminant to the number of x-intercepts of a parabola

Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: The Quadratic Formula and the Discriminant

  1. 111th Grade Factoring and Solving Quadratics Worksheet — Level 1
  2. 211th Grade Factoring and Solving Quadratics Worksheet — Level 2
  3. 311th Grade Factoring and Solving Quadratics Worksheet — Level 3
  4. 411th Grade Systems of Equations Worksheet — Level 1
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practice with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

11th Grade students (typically ages 16–17) working on quadratics. It is pitched at level 2 · standard.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try 11th Grade Factoring and Solving Quadratics Worksheet — Level 1.

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Success TutoringWorksheets11th Grade · Mathematics · Quadratics

The Quadratic Formula and the Discriminant

What you need to know

Any quadratic equation can be written as ax[sup 2] + bx + c = 0 with a ≠ 0. Factoring is fastest when it works, but the quadratic formula always works: x = [frac −b ± √(b[sup 2] − 4ac)/2a]. Identify a, b and c carefully (including their signs), substitute, and simplify. The whole of the top line is divided by 2a.

The part under the square root, Δ = b[sup 2] − 4ac, is called the discriminant. Its sign tells you how many real solutions the equation has before you solve it — and therefore how many times the parabola y = ax2 + bx + c crosses the x-axis.

DiscriminantNumber of real solutionsGraph of y = ax2 + bx + c
Δ > 0two distinct real solutionscrosses the x-axis twice
Δ = 0one repeated (equal) solution, x = touches the x-axis at the vertex
Δ < 0no real solutionsdoes not meet the x-axis
Δ > 0 and a perfect squaretwo rational solutions (it would have factored)crosses the x-axis twice

Solve 2x2 + 5x − 3 = 0

  1. a = 2, b = 5, c = −3
  2. Δ = 52 − 4(2)(−3) = 25 + 24 = 49 (a perfect square, so expect rational answers)
  3. x = =
  4. x = = [frac 1/2] or x = = −3
  5. Check: 2()2 + 5() − 3 = + − 3 = 0. Correct.

Solve x2 − 4x + 1 = 0, leaving your answer in exact form

  1. a = 1, b = −4, c = 1, so −b = 4
  2. Δ = (−4)2 − 4(1)(1) = 16 − 4 = 12
  3. x =
  4. Simplify the radical: √12 = √4 × √3 = 2√3
  5. x = = 2 ± √3

Solve 3x2 − 2x + 5 = 0

  1. Δ = (−2)2 − 4(3)(5) = 4 − 60 = −56
  2. Δ < 0, so there is no real solution. (The parabola y = 3x2 − 2x + 5 sits entirely above the x-axis.)

Find k so that x2 + kx + 9 = 0 has equal roots

  1. Equal roots means Δ = 0.
  2. Δ = k2 − 4(1)(9) = k2 − 36
  3. k2 − 36 = 0, so k2 = 36 and k = 6 or k = −6.
  4. Check k = 6: x2 + 6x + 9 = (x + 3)2, a repeated root at x = −3. Correct.

For what values of m does 2x2 − 3x + m = 0 have no real roots?

  1. No real roots means Δ < 0.
  2. Δ = (−3)2 − 4(2)(m) = 9 − 8m
  3. 9 − 8m < 0, so 8m > 9 and m > [frac 9/8].

Show that x2 + 3x + 5 is positive for all real x

  1. Δ = 32 − 4(1)(5) = 9 − 20 = −11 < 0, so the parabola never touches the x-axis.
  2. a = 1 > 0, so the parabola opens upwards and lies entirely above the x-axis.
  3. Therefore x2 + 3x + 5 > 0 for all real x. (Such a quadratic is called positive definite.)
Common mistake: sign errors with b and c. If the equation is x2 − 4x + 1 = 0 then b = −4, so −b = +4 and b2 = 16. If c is negative, −4ac becomes positive, so the discriminant gets bigger, not smaller. Another common mistake: dividing only part of the numerator by 2a. In , both the 4 and the 2√3 are divided by 2 to give 2 ± √3. Writing 4 ± √3 or 2 ± 2√3 is wrong.

Check yourself

  1. Find the discriminant of x2 + 6x + 9 = 0 and state the number of real solutions.
  2. Solve x2 + 3x − 10 = 0 using the quadratic formula.
  3. Solve x2 − 6x + 4 = 0, leaving your answer in exact form.
  4. How many real solutions does 2x2 + x + 1 = 0 have?
  5. Find k so that x2 − kx + 4 = 0 has equal roots.
  6. For what values of m does x2 + 4x + m = 0 have two distinct real roots?

Answers: 1. Δ = 36 − 36 = 0, one repeated solution (x = −3) 2. Δ = 49, x = , so x = 2 or x = −5 3. Δ = 20, x = = 3 ± √5 4. Δ = 1 − 8 = −7, so none 5. k2 − 16 = 0, so k = ±4 6. 16 − 4m > 0, so m < 4

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Last reviewed
1 October 2026
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