Question

Which among the following quadratic equation has no real solution?

• A. $$x ^ 2 + 2x - 8 = 0$$
• B. $$x ^ 2 + 2x + 8 = 0$$
• C. $$x ^ 2 + 4x + 4 = 0$$
• D. $$x ^ 2 - 4x + 4 = 0$$

Answer 1

Answer:

$$x ^ 2 + 2x + 8 = 0$$

Understanding Quadratic Equations and Real Solutions

Quadratic Equation Basics

• A quadratic equation is a polynomial equation of the second degree, generally expressed in the form ax² + bx + c = 0, where 'a', 'b', and 'c' are coefficients, and 'a' is not equal to 0.

• The solutions to a quadratic equation are also known as its roots.

The Discriminant

• The nature of the roots (real or imaginary) of a quadratic equation can be determined using the discriminant.

• The discriminant is represented by the symbol Δ (Delta) and is calculated as: Δ = b² - 4ac.

Conditions for Real Solutions

• If the discriminant (Δ) is greater than 0 (Δ > 0), the equation has two distinct real solutions.

• If the discriminant (Δ) is equal to 0 (Δ = 0), the equation has exactly one real solution (a repeated root).

• If the discriminant (Δ) is less than 0 (Δ < 0), the equation has no real solutions. Instead, it has two complex conjugate solutions.

Applying to the Given Equation ($x^2 + 2x + 8 = 0$)

• In the equation $x^2 + 2x + 8 = 0$, we have:

  • a = 1

  • b = 2

  • c = 8

• Calculate the discriminant:

  • Δ = b² - 4ac

  • Δ = (2)² - 4(1)(8)

  • Δ = 4 - 32

  • Δ = -28

• Since the discriminant (Δ) is -28, which is less than 0 (Δ < 0), this quadratic equation has no real solutions.