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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]
```
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