Answer
Here are the solutions to the equations:
q) x = 0, 1, 2
r) x = 0, 2, –4/3
s) x = –1/2, –2/3
t) x = –1, –5/3
u) x = 0, –1/6, –3/7
v) x = –1, –2
Solution
We will solve each equation one by one.
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q) Solve x³ – 3x² + 2x = 0.
Step 1. Factor out x:
x (x² – 3x + 2) = 0
Step 2. Factor the quadratic:
x² – 3x + 2 = (x – 1)(x – 2)
Thus, the equation factors as:
x (x – 1)(x – 2) = 0
Step 3. Set each factor equal to zero:
x = 0 or x – 1 = 0 or x – 2 = 0
Therefore, the solutions are:
x = 0, 1, 2
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r) Solve 3x³ – 2x² – 8x = 0.
Step 1. Factor out x:
x (3x² – 2x – 8) = 0
So, one solution is:
x = 0
Step 2. Solve the quadratic 3x² – 2x – 8 = 0.
Use the quadratic formula:
x = [–b ± √(b² – 4ac)] / (2a)
Here, a = 3, b = –2, c = –8.
Compute the discriminant:
Δ = (–2)² – 4·3·(–8) = 4 + 96 = 100
Then:
x = [2 ± 10] / (6)
This gives:
x = (2 + 10) / 6 = 12/6 = 2
x = (2 – 10) / 6 = –8/6 = –4/3
Thus, the solutions are:
x = 0, 2, –4/3
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s) Solve 6x² + 7x = –2.
Step 1. Bring all terms to one side:
6x² + 7x + 2 = 0
Step 2. Factor the quadratic. Look for two numbers that multiply to 6×2 = 12 and add to 7. They are 3 and 4.
Write:
6x² + 3x + 4x + 2 = 0
Group terms:
(6x² + 3x) + (4x + 2) = 0
Factor:
3x(2x + 1) + 2(2x + 1) = 0
Factor common term (2x + 1):
(2x + 1)(3x + 2) = 0
Set each factor equal to zero:
2x + 1 = 0 ⟹ x = –1/2
3x + 2 = 0 ⟹ x = –2/3
Thus, the solutions are:
x = –1/2, –2/3
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t) Solve 9x² + 24x + 15 = 0.
Step 1. Use the quadratic formula with a = 9, b = 24, c = 15.
Compute the discriminant:
Δ = 24² – 4·9·15 = 576 – 540 = 36
Take the square root:
√36 = 6
Then:
x = [–24 ± 6] / (2·9) = [–24 ± 6] / 18
Thus,
x = (–24 + 6)/18 = –18/18 = –1
x = (–24 – 6)/18 = –30/18 = –5/3
So, the solutions are:
x = –1, –5/3
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u) Solve 84x³ + 50x² + 6x = 0.
Step 1. Factor out the common factor 2x:
2x (42x² + 25x + 3) = 0
So one solution is:
x = 0
Step 2. Solve the quadratic 42x² + 25x + 3 = 0.
Use the quadratic formula where a = 42, b = 25, c = 3.
First, find the discriminant:
Δ = 25² – 4·42·3 = 625 – 504 = 121
√121 = 11
Now,
x = [–25 ± 11] / (2·42) = [–25 ± 11] / 84
This gives:
x = (–25 + 11) / 84 = –14/84 = –1/6
x = (–25 – 11) / 84 = –36/84 = –3/7
Thus, the solutions are:
x = 0, –1/6, –3/7
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v) Solve 2(x + 1)² = (2x + 2)(2x + 3).
Step 1. Notice that 2x + 2 = 2(x + 1). Then the right-hand side becomes:
(2(x + 1))(2x + 3)
The equation now reads:
2(x + 1)² = 2(x + 1)(2x + 3)
Step 2. Divide both sides by 2 (provided 2 ≠ 0):
(x + 1)² = (x + 1)(2x + 3)
Step 3. Subtract the right side from both sides:
(x + 1)² – (x + 1)(2x + 3) = 0
Factor out (x + 1):
(x + 1)[(x + 1) – (2x + 3)] = 0
Simplify the bracket:
(x + 1) – (2x + 3) = x + 1 – 2x – 3 = –x – 2
Thus, we have:
(x + 1)(–x – 2) = 0
Set each factor equal to zero:
Factor 1: x + 1 = 0 ⟹ x = –1
Factor 2: –x – 2 = 0 ⟹ –x = 2 ⟹ x = –2
So, the solutions are:
x = –1, –2
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Summary of Answers:
q) x = 0, 1, 2
r) x = 0, 2, –4/3
s) x = –1/2, –2/3
t) x = –1, –5/3
u) x = 0, –1/6, –3/7
v) x = –1, –2
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