Answer :
Sure! Let's solve each of these equations step-by-step to find the real solutions.
### Equation 16: [tex]\( x^3 = 27 \)[/tex]
To solve this equation, we take the cube root of both sides:
[tex]\[ x = \sqrt[3]{27} \][/tex]
The cube root of 27 is 3. Thus, the real solution is:
[tex]\[ x = 3 \][/tex]
### Equation 19: [tex]\( 64x^3 = -8 \)[/tex]
To solve for [tex]\( x \)[/tex], first divide both sides by 64:
[tex]\[ x^3 = \frac{-8}{64} \][/tex]
[tex]\[ x^3 = -\frac{1}{8} \][/tex]
Now take the cube root of both sides:
[tex]\[ x = \sqrt[3]{-\frac{1}{8}} \][/tex]
The real cube root of [tex]\(-\frac{1}{8}\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex]. So, the real solution is:
[tex]\[ x = -\frac{1}{2} \][/tex]
### Equation 22: [tex]\( x^4 - 12x^2 = 64 \)[/tex]
First, substitute [tex]\( y = x^2 \)[/tex]. The equation becomes:
[tex]\[ y^2 - 12y - 64 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1, b = -12, c = -64 \)[/tex].
This gives us:
[tex]\[ y = \frac{12 \pm \sqrt{144 + 256}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm \sqrt{400}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm 20}{2} \][/tex]
The solutions for [tex]\( y \)[/tex] are [tex]\( y = 16 \)[/tex] and [tex]\( y = -4 \)[/tex].
Now, substitute back [tex]\( y = x^2 \)[/tex].
For [tex]\( y = 16 \)[/tex], [tex]\( x^2 = 16 \)[/tex], so [tex]\( x = 4 \)[/tex] or [tex]\( x = -4 \)[/tex].
Since [tex]\( y = x^2 \)[/tex] cannot be negative, we discard [tex]\( y = -4 \)[/tex].
Thus, the real solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = -4 \][/tex]
### Equation 25: [tex]\( x^3 - 4x^2 - 7x = -10 \)[/tex]
Rearrange the equation:
[tex]\[ x^3 - 4x^2 - 7x + 10 = 0 \][/tex]
Using factoring, trial and error, or synthetic division, we determine the real roots. The real solutions are:
[tex]\[ x = -2, \quad x = 1, \quad x = 5 \][/tex]
### Equation 28: [tex]\( 6x^2 = 48x \)[/tex]
Divide both sides by 6:
[tex]\[ x^2 = 8x \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 8x = 0 \][/tex]
Factor this equation:
[tex]\[ x(x - 8) = 0 \][/tex]
The solutions are:
[tex]\[ x = 0 \quad \text{or} \quad x = 8 \][/tex]
### Equation 31: [tex]\( 4x^3 = 4x^2 + 3x \)[/tex]
Rearrange the equation:
[tex]\[ 4x^3 - 4x^2 - 3x = 0 \][/tex]
Factor out the greatest common factor, [tex]\( x \)[/tex]:
[tex]\[ x(4x^2 - 4x - 3) = 0 \][/tex]
Set each factor to zero and solve:
[tex]\[ x = 0 \][/tex]
[tex]\[ 4x^2 - 4x - 3 = 0 \][/tex]
For the quadratic:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 4, b = -4, c = -3 \)[/tex].
[tex]\[ x = \frac{4 \pm \sqrt{16 + 48}}{8} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{64}}{8} \][/tex]
[tex]\[ x = \frac{4 \pm 8}{8} \][/tex]
This gives us [tex]\( x = 1.5 \)[/tex] and [tex]\( x = -0.5 \)[/tex].
So, the real solutions are:
[tex]\[ x = 0, \quad x = 1.5, \quad x = -0.5 \][/tex]
### Equation 34: [tex]\( x^4 - 4x^3 - x^2 + 16x = 12 \)[/tex]
Rearrange the equation:
[tex]\[ x^4 - 4x^3 - x^2 + 16x - 12 = 0 \][/tex]
Using methods such as factoring or polynomial division, we find the real roots. The real solutions are:
[tex]\[ x = -2, \quad x = 1, \quad x = 2, \quad x = 3 \][/tex]
I hope this step-by-step breakdown helps! If you have any questions about these solutions, feel free to ask.
### Equation 16: [tex]\( x^3 = 27 \)[/tex]
To solve this equation, we take the cube root of both sides:
[tex]\[ x = \sqrt[3]{27} \][/tex]
The cube root of 27 is 3. Thus, the real solution is:
[tex]\[ x = 3 \][/tex]
### Equation 19: [tex]\( 64x^3 = -8 \)[/tex]
To solve for [tex]\( x \)[/tex], first divide both sides by 64:
[tex]\[ x^3 = \frac{-8}{64} \][/tex]
[tex]\[ x^3 = -\frac{1}{8} \][/tex]
Now take the cube root of both sides:
[tex]\[ x = \sqrt[3]{-\frac{1}{8}} \][/tex]
The real cube root of [tex]\(-\frac{1}{8}\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex]. So, the real solution is:
[tex]\[ x = -\frac{1}{2} \][/tex]
### Equation 22: [tex]\( x^4 - 12x^2 = 64 \)[/tex]
First, substitute [tex]\( y = x^2 \)[/tex]. The equation becomes:
[tex]\[ y^2 - 12y - 64 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1, b = -12, c = -64 \)[/tex].
This gives us:
[tex]\[ y = \frac{12 \pm \sqrt{144 + 256}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm \sqrt{400}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm 20}{2} \][/tex]
The solutions for [tex]\( y \)[/tex] are [tex]\( y = 16 \)[/tex] and [tex]\( y = -4 \)[/tex].
Now, substitute back [tex]\( y = x^2 \)[/tex].
For [tex]\( y = 16 \)[/tex], [tex]\( x^2 = 16 \)[/tex], so [tex]\( x = 4 \)[/tex] or [tex]\( x = -4 \)[/tex].
Since [tex]\( y = x^2 \)[/tex] cannot be negative, we discard [tex]\( y = -4 \)[/tex].
Thus, the real solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = -4 \][/tex]
### Equation 25: [tex]\( x^3 - 4x^2 - 7x = -10 \)[/tex]
Rearrange the equation:
[tex]\[ x^3 - 4x^2 - 7x + 10 = 0 \][/tex]
Using factoring, trial and error, or synthetic division, we determine the real roots. The real solutions are:
[tex]\[ x = -2, \quad x = 1, \quad x = 5 \][/tex]
### Equation 28: [tex]\( 6x^2 = 48x \)[/tex]
Divide both sides by 6:
[tex]\[ x^2 = 8x \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 8x = 0 \][/tex]
Factor this equation:
[tex]\[ x(x - 8) = 0 \][/tex]
The solutions are:
[tex]\[ x = 0 \quad \text{or} \quad x = 8 \][/tex]
### Equation 31: [tex]\( 4x^3 = 4x^2 + 3x \)[/tex]
Rearrange the equation:
[tex]\[ 4x^3 - 4x^2 - 3x = 0 \][/tex]
Factor out the greatest common factor, [tex]\( x \)[/tex]:
[tex]\[ x(4x^2 - 4x - 3) = 0 \][/tex]
Set each factor to zero and solve:
[tex]\[ x = 0 \][/tex]
[tex]\[ 4x^2 - 4x - 3 = 0 \][/tex]
For the quadratic:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 4, b = -4, c = -3 \)[/tex].
[tex]\[ x = \frac{4 \pm \sqrt{16 + 48}}{8} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{64}}{8} \][/tex]
[tex]\[ x = \frac{4 \pm 8}{8} \][/tex]
This gives us [tex]\( x = 1.5 \)[/tex] and [tex]\( x = -0.5 \)[/tex].
So, the real solutions are:
[tex]\[ x = 0, \quad x = 1.5, \quad x = -0.5 \][/tex]
### Equation 34: [tex]\( x^4 - 4x^3 - x^2 + 16x = 12 \)[/tex]
Rearrange the equation:
[tex]\[ x^4 - 4x^3 - x^2 + 16x - 12 = 0 \][/tex]
Using methods such as factoring or polynomial division, we find the real roots. The real solutions are:
[tex]\[ x = -2, \quad x = 1, \quad x = 2, \quad x = 3 \][/tex]
I hope this step-by-step breakdown helps! If you have any questions about these solutions, feel free to ask.
