5.5 homework – due tomorrow wed. 1/7/26
pg. 355-356 #3, 6, 10, 17, 20, 26
solve each equation. see examples 1–5.
3. ((x + 10)(x - 5) = 0) 4. ((x + 7)(x + 3) = 0) (\boldsymbol{5}). ((2k - 5)(3k + 8) = 0)
6. ((3q - 4)(2q + 5) = 0) 7. (x^2 - 3x - 10 = 0) 8. (x^2 + x - 12 = 0)
9. (x^2 + 9x + 18 = 0) 10. (x^2 - 18x + 80 = 0) (\boldsymbol{11}). (2x^2 = 7x + 4)
12. (2x^2 = 3 - x) 13. (15x^2 - 7x = 4) 14. (3x^2 + 3 = -10x)
(\boldsymbol{15}). (4p^2 + 16p = 0) 16. (2t^2 - 8t = 0) 17. (6x^2 - 36x = 0)
18. (3x^2 - 27x = 0) (\boldsymbol{19}). (4p^2 - 16 = 0) 20. (9z^2 - 81 = 0)
21. (-3x^2 + 27 = 0) 22. (-2x^2 + 8 = 0) 23. (-x^2 = 9 - 6x)
24. (-x^2 - 8x = 16) 25. (9x^2 + 24x + 16 = 0) 26. (4x^2 - 20x + 25 = 0)

Question

5.5 homework – due tomorrow wed. 1/7/26
pg. 355-356 #3, 6, 10, 17, 20, 26
solve each equation. see examples 1–5.
3. ((x + 10)(x - 5) = 0)  4. ((x + 7)(x + 3) = 0)  (\boldsymbol{5}). ((2k - 5)(3k + 8) = 0)
6. ((3q - 4)(2q + 5) = 0)  7. (x^2 - 3x - 10 = 0)  8. (x^2 + x - 12 = 0)
9. (x^2 + 9x + 18 = 0)  10. (x^2 - 18x + 80 = 0)  (\boldsymbol{11}). (2x^2 = 7x + 4)
12. (2x^2 = 3 - x)  13. (15x^2 - 7x = 4)  14. (3x^2 + 3 = -10x)
(\boldsymbol{15}). (4p^2 + 16p = 0)  16. (2t^2 - 8t = 0)  17. (6x^2 - 36x = 0)
18. (3x^2 - 27x = 0)  (\boldsymbol{19}). (4p^2 - 16 = 0)  20. (9z^2 - 81 = 0)
21. (-3x^2 + 27 = 0)  22. (-2x^2 + 8 = 0)  23. (-x^2 = 9 - 6x)
24. (-x^2 - 8x = 16)  25. (9x^2 + 24x + 16 = 0)  26. (4x^2 - 20x + 25 = 0)

Accepted Answer

To solve these equations, we'll use the **Zero-Product Property** (if $ ab = 0 $, then $ a = 0 $ or $ b = 0 $) or factoring/isolating the variable. Let’s solve a few selected problems:

### Problem 3: $ (x + 10)(x - 5) = 0 $
# Explanation:
## Step 1: Apply Zero-Product Property
If $ (x + 10)(x - 5) = 0 $, then either $ x + 10 = 0 $ or $ x - 5 = 0 $.

## Step 2: Solve for $ x $ in each case
- For $ x + 10 = 0 $: Subtract 10 from both sides.
  $ x = -10 $
- For $ x - 5 = 0 $: Add 5 to both sides.
  $ x = 5 $

# Answer: $ x = -10 $ or $ x = 5 $

### Problem 6: $ (3q - 4)(2q + 5) = 0 $
# Explanation:
## Step 1: Apply Zero-Product Property
If $ (3q - 4)(2q + 5) = 0 $, then either $ 3q - 4 = 0 $ or $ 2q + 5 = 0 $.

## Step 2: Solve for $ q $ in each case
- For $ 3q - 4 = 0 $: Add 4 to both sides, then divide by 3.
  $ 3q = 4 \implies q = \frac{4}{3} $
- For $ 2q + 5 = 0 $: Subtract 5 from both sides, then divide by 2.
  $ 2q = -5 \implies q = -\frac{5}{2} $

# Answer: $ q = \frac{4}{3} $ or $ q = -\frac{5}{2} $

### Problem 10: $ x^2 - 18x + 80 = 0 $
# Explanation:
## Step 1: Factor the quadratic
We need two numbers that multiply to $ 80 $ and add to $ -18 $. These numbers are $ -10 $ and $ -8 $.
Thus, $ x^2 - 18x + 80 = (x - 10)(x - 8) $.

## Step 2: Apply Zero-Product Property
If $ (x - 10)(x - 8) = 0 $, then either $ x - 10 = 0 $ or $ x - 8 = 0 $.

## Step 3: Solve for $ x $
- $ x - 10 = 0 \implies x = 10 $
- $ x - 8 = 0 \implies x = 8 $

# Answer: $ x = 10 $ or $ x = 8 $

### Problem 17: $ 6x^2 - 36x = 0 $
# Explanation:
## Step 1: Factor out the GCF
The greatest common factor (GCF) of $ 6x^2 $ and $ -36x $ is $ 6x $.
Factor: $ 6x(x - 6) = 0 $

## Step 2: Apply Zero-Product Property
If $ 6x(x - 6) = 0 $, then either $ 6x = 0 $ or $ x - 6 = 0 $.

## Step 3: Solve for $ x $
- $ 6x = 0 \implies x = 0 $
- $ x - 6 = 0 \implies x = 6 $

# Answer: $ x = 0 $ or $ x = 6 $

### Problem 20: $ 9z^2 - 81 = 0 $
# Explanation:
## Step 1: Factor the difference of squares
$ 9z^2 - 81 = 9(z^2 - 9) = 9(z - 3)(z + 3) $ (since $ z^2 - 9 $ is a difference of squares: $ z^2 - 3^2 = (z - 3)(z + 3) $).

## Step 2: Apply Zero-Product Property
If $ 9(z - 3)(z + 3) = 0 $, then $ z - 3 = 0 $ or $ z + 3 = 0 $ (since $ 9
eq 0 $).

## Step 3: Solve for $ z $
- $ z - 3 = 0 \implies z = 3 $
- $ z + 3 = 0 \implies z = -3 $

# Answer: $ z = 3 $ or $ z = -3 $

### Problem 26: $ 4x^2 - 20x + 25 = 0 $
# Explanation:
## Step 1: Factor the perfect square trinomial
$ 4x^2 - 20x + 25 $ is a perfect square: $ (2x - 5)^2 $ (since $ (2x)^2 = 4x^2 $, $ 2(2x)(-5) = -20x $, and $ (-5)^2 = 25 $).

## Step 2: Solve $ (2x - 5)^2 = 0 $
Take the square root of both sides: $ 2x - 5 = 0 $.

## Step 3: Solve for $ x $
Add 5 to both sides: $ 2x = 5 \implies x = \frac{5}{2} $ (double root, since the square is zero).

# Answer: $ x = \frac{5}{2} $ (or $ x = 2.5 $)

For the remaining problems, follow the same logic: use the Zero-Product Property, factor the quadratic, or isolate the variable. Let me know if you need help with a specific one!

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 3: \( (x + 10)(x - 5) = 0 \)

Step 1: Apply Zero-Product Property

Step 2: Solve for \( x \) in each case

Step 1: Apply Zero-Product Property

Step 2: Solve for \( q \) in each case

Step 1: Factor the quadratic

Step 2: Apply Zero-Product Property

Step 3: Solve for \( x \)

Answer:

Problem 6: \( (3q - 4)(2q + 5) = 0 \)