Question

Solve -4x² = 5x + 9. *a x=-1 or x=-1/4 x=-1 or x = 9/4 x= 2 or x= 3 No real solutions

Answer 1

Answer:

No real solutions

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

Algebra I

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

Algebra II

• Imaginary Numbers: √-1 = i

Step-by-step explanation:

Step 1: Define

-4x² = 5x + 9

Step 2: Rewrite

Find standard form.

1. Subtract 5x on both sides:                    -4x² - 5x = 9
2. Subtract 9 on both sides:                      -4x² - 5x - 9 = 0

Step 3: Identify Variables

a = -4

b = -5

c = -9

Step 4: Solve for x

1. Substitute [QF]:                    $x=\frac{5\pm\sqrt{(-5)^2-4(-4)(-9)} }{2(-4)}$
2. Exponents:                           $x=\frac{5\pm\sqrt{25-4(-4)(-9)} }{2(-4)}$
3. Multiply:                                $x=\frac{5\pm\sqrt{25-144} }{-8}$
4. Subtract:                               $x=\frac{5\pm\sqrt{-119} }{-8}$
5. Factor:                                   $x=\frac{5\pm\sqrt{-1} \sqrt{119} }{-8}$
6. Simplify:                                $x=\frac{5\pm i\sqrt{119} }{-8}$

Here we see that we get imaginary numbers.

∴ the quadratic would have no real roots.