41. solve the equation.
64x³ + 1 = 0
42. solve the equation.
20x⁴ − 53x² + 18 = 0
43. solve the equation.
8x⁶ + 999x³ = 125
44. solve the equation.
4x⁴ − 4x² − x² + 1 = 0
45. solve the equation.
x⁶ − 9x⁴ − x² + 9 = 0

Question

Accepted Answer

### Problem 41: Solve $ 64x^3 + 1 = 0 $
# Explanation:
## Step 1: Rewrite the equation
Rewrite the equation as a sum of cubes. Recall that $ a^3 + b^3=(a + b)(a^2 - ab + b^2) $. Here, $ 64x^3=(4x)^3 $ and $ 1 = 1^3 $, so the equation becomes $ (4x)^3+1^3 = 0 $.
## Step 2: Factor using sum of cubes
Using the sum of cubes formula $ a^3 + b^3=(a + b)(a^2 - ab + b^2) $ with $ a = 4x $ and $ b = 1 $, we get:
$ (4x + 1)((4x)^2-4x\times1 + 1^2)=0 $
Simplify the quadratic factor: $ (4x + 1)(16x^2-4x + 1)=0 $
## Step 3: Solve for $ x $
Set each factor equal to zero:
- For $ 4x+1 = 0 $, we have $ 4x=-1 $, so $ x =-\frac{1}{4} $.
- For $ 16x^2-4x + 1 = 0 $, calculate the discriminant $ \Delta=b^2 - 4ac=(-4)^2-4\times16\times1=16 - 64=-48<0 $. Since the discriminant is negative, there are no real solutions from this factor, and the complex solutions are given by $ x=\frac{4\pm\sqrt{-48}}{2\times16}=\frac{4\pm4i\sqrt{3}}{32}=\frac{1\pm i\sqrt{3}}{8} $. But if we are looking for real solutions, the only real solution is $ x =-\frac{1}{4} $.

# Answer:
The real solution is $ \boldsymbol{x =-\frac{1}{4}} $, and the complex solutions are $ \boldsymbol{x=\frac{1\pm i\sqrt{3}}{8}} $ (if complex solutions are considered). If only real solutions, then $ \boldsymbol{x =-\frac{1}{4}} $.

### Problem 42: Solve $ 20x^4 - 53x^2 + 18 = 0 $
# Explanation:
## Step 1: Let $ y = x^2 $
Substitute $ y=x^2 $ (so $ y\geq0 $ for real $ x $) into the equation. The equation becomes a quadratic in $ y $: $ 20y^2-53y + 18 = 0 $.
## Step 2: Solve the quadratic equation for $ y $
Use the quadratic formula $ y=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $, where $ a = 20 $, $ b=-53 $, $ c = 18 $.
First, calculate the discriminant: $ \Delta=(-53)^2-4\times20\times18=2809 - 1440 = 1369 = 37^2 $
Then, $ y=\frac{53\pm37}{40} $
- For the plus sign: $ y=\frac{53 + 37}{40}=\frac{90}{40}=\frac{9}{4} $
- For the minus sign: $ y=\frac{53-37}{40}=\frac{16}{40}=\frac{2}{5} $
## Step 3: Solve for $ x $
Since $ y = x^2 $, we solve for $ x $ in each case:
- When $ y=\frac{9}{4} $, $ x^2=\frac{9}{4} $, so $ x=\pm\frac{3}{2} $
- When $ y=\frac{2}{5} $, $ x^2=\frac{2}{5} $, so $ x=\pm\sqrt{\frac{2}{5}}=\pm\frac{\sqrt{10}}{5} $

# Answer:
The solutions are $ \boldsymbol{x=\pm\frac{3}{2},\pm\frac{\sqrt{10}}{5}} $

### Problem 43: Solve $ 8x^6 + 999x^3 = 125 $
# Explanation:
## Step 1: Rewrite the equation
Rewrite the equation as $ 8x^6+999x^3 - 125 = 0 $. Let $ y = x^3 $, then the equation becomes a quadratic in $ y $: $ 8y^2+999y - 125 = 0 $
## Step 2: Solve the quadratic equation for $ y $
Use the quadratic formula $ y=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $, where $ a = 8 $, $ b = 999 $, $ c=-125 $
Calculate the discriminant: $ \Delta=999^2-4\times8\times(-125)=998001 + 4000 = 1002001 $
Notice that $ 1002001 = 1001^2 $ (since $ 1001^2=(1000 + 1)^2=1000000+2000 + 1 = 1002001 $)
Then, $ y=\frac{-999\pm1001}{16} $
- For the plus sign: $ y=\frac{-999 + 1001}{16}=\frac{2}{16}=\frac{1}{8} $
- For the minus sign: $ y=\frac{-999-1001}{16}=\frac{-2000}{16}=-125 $
## Step 3: Solve for $ x $
Since $ y = x^3 $, we solve for $ x $ in each case:
- When $ y=\frac{1}{8} $, $ x^3=\frac{1}{8} $, so $ x=\frac{1}{2} $ (real solution, and two complex cube roots)
- When $ y=-125 $, $ x^3=-125 $, so $ x=-5 $ (real solution, and two complex cube roots)

If we consider only real solutions, the real solutions are $ x=\frac{1}{2},-5 $. If complex solutions are considered, we can find the complex roots using the cube - root formula, but for simplicity, the real solutions are $ \boldsymbol{x=\frac{1}{2},-5} $

# Answer:
The real solutions are $ \boldsymbol{x=\frac{1}{2},-5} $ (and there are complex solutions for each cubic equation)

### Problem 44: Solve $ 4x^4 - 4x^2 - x^2 + 1 = 0 $
# Explanation:
## Step 1: Simplify and factor by grouping
First, combine like terms: $ 4x^4-5x^2 + 1 = 0 $. Then, factor by grouping. Let's rewrite it as $ 4x^4-4x^2-x^2 + 1 = 0 $ (which is the original form). Group the first two and last two terms:
$ (4x^4-4x^2)-(x^2 - 1)=0 $
Factor out $ 4x^2 $ from the first group and factor the second group as a difference of squares:
$ 4x^2(x^2 - 1)-(x^2 - 1)=0 $
Now, factor out $ (x^2 - 1) $:
$ (x^2 - 1)(4x^2 - 1)=0 $
Factor $ (x^2 - 1) $ as a difference of squares $ (x - 1)(x + 1) $ and $ (4x^2 - 1) $ as a difference of squares $ (2x - 1)(2x + 1) $
So the equation becomes $ (x - 1)(x + 1)(2x - 1)(2x + 1)=0 $
## Step 2: Solve for $ x $
Set each factor equal to zero:
- $ x - 1 = 0\Rightarrow x = 1 $
- $ x + 1 = 0\Rightarrow x=-1 $
- $ 2x - 1 = 0\Rightarrow x=\frac{1}{2} $
- $ 2x + 1 = 0\Rightarrow x=-\frac{1}{2} $

# Answer:
The solutions are $ \boldsymbol{x=\pm1,\pm\frac{1}{2}} $

### Problem 45: Solve $ x^6 - 9x^4 - x^2 + 9 = 0 $
# Explanation:
## Step 1: Factor by grouping
Group the first two and last two terms:
$ (x^6 - 9x^4)-(x^2 - 9)=0 $
Factor out $ x^4 $ from the first group and factor the second group as a difference of squares:
$ x^4(x^2 - 9)-(x^2 - 9)=0 $
Now, factor out $ (x^2 - 9) $:
$ (x^2 - 9)(x^4 - 1)=0 $
Factor $ (x^2 - 9) $ as a difference of squares $ (x - 3)(x + 3) $ and $ (x^4 - 1) $ as a difference of squares $ (x^2 - 1)(x^2 + 1) $, and then factor $ (x^2 - 1) $ as $ (x - 1)(x + 1) $
So the equation becomes $ (x - 3)(x + 3)(x - 1)(x + 1)(x^2 + 1)=0 $
## Step 2: Solve for $ x $
Set each factor equal to zero:
- $ x - 3 = 0\Rightarrow x = 3 $
- $ x + 3 = 0\Rightarrow x=-3 $
- $ x - 1 = 0\Rightarrow x = 1 $
- $ x + 1 = 0\Rightarrow x=-1 $
- $ x^2+1 = 0\Rightarrow x^2=-1\Rightarrow x=\pm i $ (complex solutions)

If we consider only real solutions, the real solutions are $ x=\pm1,\pm3 $. If complex solutions are considered, we include $ x=\pm i $ as well.

# Answer:
The real solutions are $ \boldsymbol{x=\pm1,\pm3} $, and the complex solutions are $ \boldsymbol{x=\pm i} $ (if complex solutions are considered). If only real solutions, then $ \boldsymbol{x=\pm1,\pm3} $

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QUESTION IMAGE

Question

Explanation:

Problem 41: Solve \( 64x^3 + 1 = 0 \)

Step 1: Rewrite the equation

Step 2: Factor using sum of cubes

Step 3: Solve for \( x \)

Step 1: Let \( y = x^2 \)

Step 2: Solve the quadratic equation for \( y \)

Step 3: Solve for \( x \)

Step 1: Rewrite the equation

Step 2: Solve the quadratic equation for \( y \)

Step 3: Solve for \( x \)

Answer:

Problem 42: Solve \( 20x^4 - 53x^2 + 18 = 0 \)