Question

Solve the equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.

Answer 1

Answer:

No real roots.

General Formulas and Concepts:

Pre-Algebra

• Order of Operations: BPEMDAS

Algebra I

• Standard Form: ax² + bx + c = 0
• Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Algebra II

• Imaginary Roots: √-1 = i

Step-by-step explanation:

Step 1: Define

-3x² + 2x = 1

Step 2: Rewrite in Standard Form

1. Subtract 1 on both sides:                    -3x² + 2x - 1 = 0

Step 3: Define

a = -3

b = 2

c = -1

Step 4: Find roots

1. Substitute in variables:                    $x=\frac{-2\pm\sqrt{2^2-4(-3)(-1)} }{2(-3)}$
2. Exponents:                                       $x=\frac{-2\pm\sqrt{4-4(-3)(-1)} }{2(-3)}$
3. Multiply:                                            $x=\frac{-2\pm\sqrt{4-12} }{-6}$
4. Subtract:                                           $x=\frac{-2\pm\sqrt{-8} }{-6}$

Here we see that we cannot take the square root of a negative number. We will get no real roots and only imaginary ones.