There’s no solution because x is less than 3 there is no way is greater than 5.
Hope this helps!
Answer:
B. x = -1 ± i
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Algebra I</u>
- Standard Form: ax² + bx + c = 0
- Factoring
- Quadratic Formula:

<u>Algebra II</u>
- Imaginary Numbers: √-1 = i
Step-by-step explanation:
<u>Step 1: Define</u>
x² + 2x = -2
<u>Step 2: Identify Variables</u>
- Rewrite Quadratic in Standard Form [Addition Property of Equality]: x² + 2x + 2 = 0
- Break up Quadratic: a = 1, b = 2, c = 2
<u>Step 3: Solve for </u><em><u>x</u></em>
- Substitute in variables [Quadratic Formula]:

- [√Radical] Evaluate exponents:

- Multiply:

- [√Radical] Subtract:

- [√Radical] Factor:

- [√Radicals] Simplify:

- Factor:

- Divide:

No, a cubic equation can not have three complex roots. This is because it turns twice and one end goes to positive infinity and one end goes to negative infinity. Thus, one of these MUST cross the x-axis at some point, meaning y = 0 and a real root exists.
Yes, a cubic equation can have three real roots if it cuts the x-axis three times.