Math Problem Statement

how to solve this?

How to solve the quadratic equation x^2 + 4x + 5 = 0?

Solution

The equation you are solving in the image is a quadratic equation:

x2+4x+5=0x^2 + 4x + 5 = 0

You correctly identified the coefficients:

  • a=1a = 1
  • b=4b = 4
  • c=5c = 5

The next step is to apply the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values a=1a = 1, b=4b = 4, and c=5c = 5 into the formula:

x=4±424(1)(5)2(1)x = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)}

This simplifies to:

x=4±16202x = \frac{-4 \pm \sqrt{16 - 20}}{2} x=4±42x = \frac{-4 \pm \sqrt{-4}}{2}

Now, you encounter a square root of a negative number, 4\sqrt{-4}, which implies that the solutions are complex numbers. In complex numbers, 1=i\sqrt{-1} = i, where ii is the imaginary unit. Therefore:

4=2i\sqrt{-4} = 2i

Now substitute this back into the formula:

x=4±2i2x = \frac{-4 \pm 2i}{2}

Simplify:

x=2±ix = -2 \pm i

Thus, the two complex solutions are:

x=2+iorx=2ix = -2 + i \quad \text{or} \quad x = -2 - i

These are the final solutions to the quadratic equation.


Would you like any further explanation or details on this?

Here are five related questions that expand upon this topic:

  1. What are the conditions for a quadratic equation to have real vs complex solutions?
  2. How do you graph quadratic equations with complex solutions?
  3. Can the discriminant (b24acb^2 - 4ac) predict the nature of the roots?
  4. How do imaginary numbers like ii behave in calculations involving powers?
  5. What are the applications of complex numbers in physics or engineering?

Tip: The discriminant (b24acb^2 - 4ac) is key to knowing whether your quadratic has real or complex solutions.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Complex Numbers

Formulas

Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic formula
Discriminant for determining real or complex roots

Suitable Grade Level

Grades 9-12