The Quadratic Formula

Using the quadratic formula and the discriminant

The Quadratic Formula

Standard Form

A quadratic equation in standard form: ax2+bx+c=0ax^2 + bx + c = 0

where a≠0a \neq 0

The Quadratic Formula

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula gives the solutions to any quadratic equation.

The Discriminant

The expression under the square root is called the discriminant: Δ=b2−4ac\Delta = b^2 - 4ac

The discriminant tells us about the solutions:

  • Δ>0\Delta > 0: Two real solutions
  • Δ=0\Delta = 0: One real solution (repeated root)
  • Δ<0\Delta < 0: No real solutions (two complex solutions)

Steps to Use the Formula

  1. Write the equation in standard form
  2. Identify aa, bb, and cc
  3. Substitute into the formula
  4. Simplify the result

📚 Practice Problems

1Problem 1easy

❓ Question:

Use the discriminant to determine the number of real solutions: x2−6x+9=0x^2 - 6x + 9 = 0

💡 Show Solution

Identify: a=1a = 1, b=−6b = -6, c=9c = 9

Calculate the discriminant: Δ=b2−4ac\Delta = b^2 - 4ac =(−6)2−4(1)(9)= (-6)^2 - 4(1)(9) =36−36= 36 - 36 =0= 0

Since Δ=0\Delta = 0, there is one real solution (a repeated root).

Answer: One real solution

2Problem 2medium

❓ Question:

Solve using the quadratic formula: x2−4x−5=0x^2 - 4x - 5 = 0

💡 Show Solution

Identify: a=1a = 1, b=−4b = -4, c=−5c = -5

Substitute into the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=−(−4)±(−4)2−4(1)(−5)2(1)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}

x=4±16+202x = \frac{4 \pm \sqrt{16 + 20}}{2}

x=4±362x = \frac{4 \pm \sqrt{36}}{2}

x=4±62x = \frac{4 \pm 6}{2}

Two solutions: x=4+62=102=5x = \frac{4 + 6}{2} = \frac{10}{2} = 5 x=4−62=−22=−1x = \frac{4 - 6}{2} = \frac{-2}{2} = -1

Answer: x=5x = 5 or x=−1x = -1

3Problem 3hard

❓ Question:

Solve: 2x2+3x−1=02x^2 + 3x - 1 = 0

💡 Show Solution

Identify: a=2a = 2, b=3b = 3, c=−1c = -1

Substitute: x=−3±(3)2−4(2)(−1)2(2)x = \frac{-3 \pm \sqrt{(3)^2 - 4(2)(-1)}}{2(2)}

x=−3±9+84x = \frac{-3 \pm \sqrt{9 + 8}}{4}

x=−3±174x = \frac{-3 \pm \sqrt{17}}{4}

This cannot be simplified further.

Answer: x=−3+174x = \frac{-3 + \sqrt{17}}{4} or x=−3−174x = \frac{-3 - \sqrt{17}}{4}

(Approximately: x≈0.28x \approx 0.28 or x≈−1.78x \approx -1.78)

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics