At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To solve the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex], follow these detailed steps:
1. Identify the type of equation:
This is a polynomial equation of degree 6.
2. Rewrite the equation:
Introduce a substitution to simplify the polynomial. Let [tex]\( y = x^3 \)[/tex]. Thus the equation becomes:
[tex]\[ y^2 - y - 2 = 0 \][/tex]
3. Solve the quadratic equation:
The equation [tex]\( y^2 - y - 2 = 0 \)[/tex] is a standard quadratic equation and can be solved using the quadratic formula. Recall that for a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex], the solutions are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex]. Plugging these values in:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{2} \][/tex]
Thus, the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{1 + 3}{2} = 2 \quad \text{and} \quad y = \frac{1 - 3}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = x^3 \)[/tex]:
We now have two equations to solve for [tex]\( x \)[/tex]:
[tex]\[ x^3 = 2 \quad \text{and} \quad x^3 = -1 \][/tex]
5. Solve [tex]\( x^3 = 2 \)[/tex]:
The solutions to [tex]\( x^3 = 2 \)[/tex] are the cube roots of 2:
[tex]\[ x = 2^{1/3}, \; \omega 2^{1/3}, \; \omega^2 2^{1/3} \][/tex]
where [tex]\( \omega \)[/tex] is a primitive cube root of unity satisfying [tex]\( \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)[/tex] and [tex]\( \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)[/tex].
Therefore:
[tex]\[ x = 2^{1/3}, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
6. Solve [tex]\( x^3 = -1 \)[/tex]:
The solutions to [tex]\( x^3 = -1 \)[/tex] are the cube roots of -1:
[tex]\[ x = -1, \; \omega, \; \omega^2 \][/tex]
Note here [tex]\( \omega \)[/tex] and [tex]\( \omega^2 \)[/tex] remain the same as defined previously since they are roots of unity.
Therefore:
[tex]\[ x = -1, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i \][/tex]
7. Compile all solutions:
Collecting all solutions from steps 5 and 6, we get:
[tex]\[ x = -1, 2^{1/3}, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
Thus, the solutions to the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex] are:
[tex]\[ -1, \; 2^{1/3}, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
1. Identify the type of equation:
This is a polynomial equation of degree 6.
2. Rewrite the equation:
Introduce a substitution to simplify the polynomial. Let [tex]\( y = x^3 \)[/tex]. Thus the equation becomes:
[tex]\[ y^2 - y - 2 = 0 \][/tex]
3. Solve the quadratic equation:
The equation [tex]\( y^2 - y - 2 = 0 \)[/tex] is a standard quadratic equation and can be solved using the quadratic formula. Recall that for a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex], the solutions are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex]. Plugging these values in:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{2} \][/tex]
Thus, the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{1 + 3}{2} = 2 \quad \text{and} \quad y = \frac{1 - 3}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = x^3 \)[/tex]:
We now have two equations to solve for [tex]\( x \)[/tex]:
[tex]\[ x^3 = 2 \quad \text{and} \quad x^3 = -1 \][/tex]
5. Solve [tex]\( x^3 = 2 \)[/tex]:
The solutions to [tex]\( x^3 = 2 \)[/tex] are the cube roots of 2:
[tex]\[ x = 2^{1/3}, \; \omega 2^{1/3}, \; \omega^2 2^{1/3} \][/tex]
where [tex]\( \omega \)[/tex] is a primitive cube root of unity satisfying [tex]\( \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)[/tex] and [tex]\( \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)[/tex].
Therefore:
[tex]\[ x = 2^{1/3}, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
6. Solve [tex]\( x^3 = -1 \)[/tex]:
The solutions to [tex]\( x^3 = -1 \)[/tex] are the cube roots of -1:
[tex]\[ x = -1, \; \omega, \; \omega^2 \][/tex]
Note here [tex]\( \omega \)[/tex] and [tex]\( \omega^2 \)[/tex] remain the same as defined previously since they are roots of unity.
Therefore:
[tex]\[ x = -1, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i \][/tex]
7. Compile all solutions:
Collecting all solutions from steps 5 and 6, we get:
[tex]\[ x = -1, 2^{1/3}, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
Thus, the solutions to the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex] are:
[tex]\[ -1, \; 2^{1/3}, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
