Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To solve the equation \( x^4 + 6x^2 + 5 = 0 \) using substitution, we'll follow these steps:
1. Substitute Variables: First, let's make a substitution \( u = x^2 \). This allows us to transform the given equation into a simpler quadratic form.
2. Rewrite the Equation: Substitute \( u = x^2 \) into the original equation:
[tex]\[ x^4 + 6x^2 + 5 = 0 \implies u^2 + 6u + 5 = 0 \][/tex]
3. Solve the Quadratic Equation: Next, we solve the quadratic equation \( u^2 + 6u + 5 = 0 \) for \( u \). This can be done using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 6 \), and \( c = 5 \).
[tex]\[ u = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-6 \pm \sqrt{36 - 20}}{2} \][/tex]
[tex]\[ u = \frac{-6 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ u = \frac{-6 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ u = \frac{-6 + 4}{2} = \frac{-2}{2} = -1 \quad \text{and} \quad u = \frac{-6 - 4}{2} = \frac{-10}{2} = -5 \][/tex]
4. Return to Original Variables: Now, reverse the substitution to return to the variable \( x \). We have \( u = x^2 \), so we need to solve the equations \( x^2 = -1 \) and \( x^2 = -5 \) for \( x \).
- For \( x^2 = -1 \):
[tex]\[ x = \pm \sqrt{-1} = \pm i \][/tex]
- For \( x^2 = -5 \):
[tex]\[ x = \pm \sqrt{-5} = \pm i\sqrt{5} \][/tex]
5. Collect all Solutions: Combining these, the solutions to the original equation are:
[tex]\[ x = \pm i \quad \text{and} \quad x = \pm i\sqrt{5} \][/tex]
So, the solutions of the equation \( x^4 + 6x^2 + 5 = 0 \) are \( x = \pm i \) and \( x = \pm i\sqrt{5} \). Thus, the correct answer from the given options is:
```
[tex]$x= \pm i$[/tex] and [tex]$x= \pm i \sqrt{5}$[/tex]
```
1. Substitute Variables: First, let's make a substitution \( u = x^2 \). This allows us to transform the given equation into a simpler quadratic form.
2. Rewrite the Equation: Substitute \( u = x^2 \) into the original equation:
[tex]\[ x^4 + 6x^2 + 5 = 0 \implies u^2 + 6u + 5 = 0 \][/tex]
3. Solve the Quadratic Equation: Next, we solve the quadratic equation \( u^2 + 6u + 5 = 0 \) for \( u \). This can be done using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 6 \), and \( c = 5 \).
[tex]\[ u = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-6 \pm \sqrt{36 - 20}}{2} \][/tex]
[tex]\[ u = \frac{-6 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ u = \frac{-6 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ u = \frac{-6 + 4}{2} = \frac{-2}{2} = -1 \quad \text{and} \quad u = \frac{-6 - 4}{2} = \frac{-10}{2} = -5 \][/tex]
4. Return to Original Variables: Now, reverse the substitution to return to the variable \( x \). We have \( u = x^2 \), so we need to solve the equations \( x^2 = -1 \) and \( x^2 = -5 \) for \( x \).
- For \( x^2 = -1 \):
[tex]\[ x = \pm \sqrt{-1} = \pm i \][/tex]
- For \( x^2 = -5 \):
[tex]\[ x = \pm \sqrt{-5} = \pm i\sqrt{5} \][/tex]
5. Collect all Solutions: Combining these, the solutions to the original equation are:
[tex]\[ x = \pm i \quad \text{and} \quad x = \pm i\sqrt{5} \][/tex]
So, the solutions of the equation \( x^4 + 6x^2 + 5 = 0 \) are \( x = \pm i \) and \( x = \pm i\sqrt{5} \). Thus, the correct answer from the given options is:
```
[tex]$x= \pm i$[/tex] and [tex]$x= \pm i \sqrt{5}$[/tex]
```
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.