3. $x^2 - 7x = 12$
4. $x^2 = -x + 6$
5. $p(x) = x^2 - 3x - x$
6. $9x^2 - 3x = 0$
7. $3x^2 - 20x - 7 = 0$
8. $(4x^2 - 9)(4x^2 - 5x) = 0$

Question

Accepted Answer

### Problem 3: $ x^2 - 7x = 12 $
# Explanation:
## Step 1: Rewrite in standard form
Subtract 12 from both sides to get $ x^2 - 7x - 12 = 0 $.
## Step 2: Use quadratic formula
For $ ax^2 + bx + c = 0 $, $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $. Here, $ a = 1 $, $ b = -7 $, $ c = -12 $.
Discriminant: $ (-7)^2 - 4(1)(-12) = 49 + 48 = 97 $.
So $ x = \frac{7 \pm \sqrt{97}}{2} $.
# Answer: $ x = \frac{7 + \sqrt{97}}{2}, \frac{7 - \sqrt{97}}{2} $

### Problem 4: $ x^2 = -x + 6 $
# Explanation:
## Step 1: Rewrite in standard form
$ x^2 + x - 6 = 0 $.
## Step 2: Factor the quadratic
Find two numbers that multiply to -6 and add to 1: 3 and -2.
So $ (x + 3)(x - 2) = 0 $.
## Step 3: Solve for x
$ x + 3 = 0 $ or $ x - 2 = 0 $, so $ x = -3 $ or $ x = 2 $.
# Answer: $ x = -3, 2 $

### Problem 5: $ p(x) = x^2 - 3x - x $
# Explanation:
## Step 1: Combine like terms
$ -3x - x = -4x $, so $ p(x) = x^2 - 4x $.
## Step 2: Factor (optional, for roots)
$ p(x) = x(x - 4) $. Roots at $ x = 0 $ and $ x = 4 $.
# Answer: $ p(x) = x^2 - 4x $ (or roots $ x = 0, 4 $)

### Problem 6: $ 9x^2 - 3x = 0 $
# Explanation:
## Step 1: Factor out GCF
GCF of $ 9x^2 $ and $ -3x $ is $ 3x $, so $ 3x(3x - 1) = 0 $.
## Step 2: Solve for x
$ 3x = 0 $ or $ 3x - 1 = 0 $, so $ x = 0 $ or $ x = \frac{1}{3} $.
# Answer: $ x = 0, \frac{1}{3} $

### Problem 7: $ 3x^2 - 20x - 7 = 0 $
# Explanation:
## Step 1: Factor the quadratic
Find two numbers: $ 3 \times (-7) = -21 $, numbers that multiply to -21 and add to -20: -21 and 1.
Rewrite middle term: $ 3x^2 - 21x + x - 7 = 0 $.
Factor by grouping: $ 3x(x - 7) + 1(x - 7) = 0 $, so $ (3x + 1)(x - 7) = 0 $.
## Step 2: Solve for x
$ 3x + 1 = 0 $ or $ x - 7 = 0 $, so $ x = -\frac{1}{3} $ or $ x = 7 $.
# Answer: $ x = -\frac{1}{3}, 7 $

### Problem 8: $ (4x^2 - 9)(4x^2 - 5x) = 0 $
# Explanation:
## Step 1: Set each factor to zero
First factor: $ 4x^2 - 9 = 0 $ (difference of squares: $ (2x - 3)(2x + 3) = 0 $, so $ x = \frac{3}{2}, -\frac{3}{2} $).
Second factor: $ 4x^2 - 5x = 0 $, factor out $ x $: $ x(4x - 5) = 0 $, so $ x = 0 $ or $ x = \frac{5}{4} $.
# Answer: $ x = -\frac{3}{2}, 0, \frac{3}{2}, \frac{5}{4} $

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QUESTION IMAGE

Question

Explanation:

Problem 3: \( x^2 - 7x = 12 \)

Step 1: Rewrite in standard form

Step 2: Use quadratic formula

Step 1: Rewrite in standard form

Step 2: Factor the quadratic

Step 3: Solve for x

Step 1: Combine like terms

Step 2: Factor (optional, for roots)

Answer:

Problem 4: \( x^2 = -x + 6 \)