A quadratic equation can be written ax squared + bx + c = 0 with a nonzero. Factoring, square roots, completing the square, and the quadratic formula are valid approaches; the discriminant tells how many distinct real roots to expect.

A reliable method

  1. Move all terms to one side and identify a, b, and c.
  2. Compute b squared minus 4ac.
  3. Use a convenient method; the quadratic formula works for every quadratic.
  4. Check roots in the original equation and distinguish real from nonreal solutions.

Example 1

Solve over the real numbers: \(x^{2} - 5 x + 6=0\).

  1. Move every term to one side and collect: \(x^{2} - 5 x + 6=0\).
  2. Identify \(a=1\), \(b=-5\), \(c=6\).
  3. Compute the discriminant: \(D=b^2-4ac=1\).
  4. Use \(x=\frac{-b\pm\sqrt{D}}{2a}\) and simplify.
  5. The real solution set is \(\left\{2, 3\right\}\). Substituting each solution into the original equation verifies it.

Answer: \(\left\{2, 3\right\}\)

Example 2

Solve over the real numbers: \(2 x^{2} + 3 x - 2=0\).

  1. Move every term to one side and collect: \(2 x^{2} + 3 x - 2=0\).
  2. Identify \(a=2\), \(b=3\), \(c=-2\).
  3. Compute the discriminant: \(D=b^2-4ac=25\).
  4. Use \(x=\frac{-b\pm\sqrt{D}}{2a}\) and simplify.
  5. The real solution set is \(\left\{-2, \frac{1}{2}\right\}\). Substituting each solution into the original equation verifies it.

Answer: \(\left\{-2, \frac{1}{2}\right\}\)

Common mistakes

  • The denominator 2a divides the entire formula numerator.
  • A zero discriminant gives one repeated real root, not two distinct roots.

Practice without the answer first

  1. Solve over the real numbers: \(x^{2} - 6 x + 9=0\).
  2. Solve over the real numbers: \(x^{2} + 4=0\).
  3. Solve over the real numbers: \(3 x^{2} - 2 x=4\).
Answers and step-by-step explanations

Practice 1

  1. Move every term to one side and collect: \(x^{2} - 6 x + 9=0\).
  2. Identify \(a=1\), \(b=-6\), \(c=9\).
  3. Compute the discriminant: \(D=b^2-4ac=0\).
  4. Use \(x=\frac{-b\pm\sqrt{D}}{2a}\) and simplify.
  5. The real solution set is \(\left\{3\right\}\). Substituting each solution into the original equation verifies it.

Answer: \(\left\{3\right\}\)

Practice 2

  1. Move every term to one side and collect: \(x^{2} + 4=0\).
  2. Identify \(a=1\), \(b=0\), \(c=4\).
  3. Compute the discriminant: \(D=b^2-4ac=-16\).
  4. The discriminant is negative, so there are no real solutions.

Answer: \(\emptyset\)

Practice 3

  1. Move every term to one side and collect: \(3 x^{2} - 2 x - 4=0\).
  2. Identify \(a=3\), \(b=-2\), \(c=-4\).
  3. Compute the discriminant: \(D=b^2-4ac=52\).
  4. Use \(x=\frac{-b\pm\sqrt{D}}{2a}\) and simplify.
  5. The real solution set is \(\left\{\frac{1}{3} - \frac{\sqrt{13}}{3}, \frac{1}{3} + \frac{\sqrt{13}}{3}\right\}\). Substituting each solution into the original equation verifies it.

Answer: \(\left\{\frac{1}{3} - \frac{\sqrt{13}}{3}, \frac{1}{3} + \frac{\sqrt{13}}{3}\right\}\)

Study context: This shared lesson supports math study across exams. Use it to repair a prerequisite or practice the matching skill; it is not an official SAT question set or a score prediction.

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