Answer :
Certainly! Let's find the roots of each equation step-by-step.
### a. [tex]\( x^2 + x + 1 = 0 \)[/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( x^2 + x + 1 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex].
Plugging in these values, we get:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 - 4}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{-3}}{2} \][/tex]
Since [tex]\(\sqrt{-3} = \sqrt{3}i\)[/tex], we get:
[tex]\[ x = \frac{-1 \pm \sqrt{3}i}{2} \][/tex]
Therefore, the roots are:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{3}i}{2} \quad \text{and} \quad x = -\frac{1}{2} - \frac{\sqrt{3}i}{2} \][/tex]
### b. [tex]\( x^2 + x + 3 = 0 \)[/tex]
Using the quadratic formula for [tex]\( x^2 + x + 3 = 0 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{-11}}{2} \][/tex]
Since [tex]\(\sqrt{-11} = \sqrt{11}i\)[/tex], we get:
[tex]\[ x = \frac{-1 \pm \sqrt{11}i}{2} \][/tex]
Therefore, the roots are:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{11}i}{2} \quad \text{and} \quad x = -\frac{1}{2} - \frac{\sqrt{11}i}{2} \][/tex]
### c. [tex]\( 2x^2 + 7x + 1 = 0 \)[/tex]
Using the quadratic formula for [tex]\( 2x^2 + 7x + 1 = 0 \)[/tex]:
[tex]\[ x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-7 \pm \sqrt{49 - 8}}{4} \][/tex]
[tex]\[ x = \frac{-7 \pm \sqrt{41}}{4} \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-7 + \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-7 - \sqrt{41}}{4} \][/tex]
### d. [tex]\( x^4 - 1 = 0 \)[/tex]
This can be factored as:
[tex]\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1) \][/tex]
Solving each factor separately, we get:
[tex]\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \][/tex]
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
For [tex]\( x^2 + 1 = 0 \)[/tex]:
[tex]\[ x^2 + 1 = 0 \quad \Rightarrow \quad x^2 = -1 \quad \Rightarrow \quad x = \pm i \][/tex]
So, the roots are:
[tex]\[ x = 1, -1, i, -i \][/tex]
### e. [tex]\( x^2 + 9 = 0 \)[/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 9 = 0 \quad \Rightarrow \quad x^2 = -9 \quad \Rightarrow \quad x = \pm \sqrt{-9} \quad \Rightarrow \quad x = \pm 3i \][/tex]
So, the roots are:
[tex]\[ x = 3i \quad \text{and} \quad x = -3i \][/tex]
In summary:
a. [tex]\( x = -\frac{1}{2} + \frac{\sqrt{3}i}{2} \)[/tex] and [tex]\( x = -\frac{1}{2} - \frac{\sqrt{3}i}{2} \)[/tex]
b. [tex]\( x = -\frac{1}{2} + \frac{\sqrt{11}i}{2} \)[/tex] and [tex]\( x = -\frac{1}{2} - \frac{\sqrt{11}i}{2} \)[/tex]
c. [tex]\( x = \frac{-7 + \sqrt{41}}{4} \)[/tex] and [tex]\( x = \frac{-7 - \sqrt{41}}{4} \)[/tex]
d. [tex]\( x = 1, -1, i, -i \)[/tex]
e. [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex]
### a. [tex]\( x^2 + x + 1 = 0 \)[/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( x^2 + x + 1 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex].
Plugging in these values, we get:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 - 4}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{-3}}{2} \][/tex]
Since [tex]\(\sqrt{-3} = \sqrt{3}i\)[/tex], we get:
[tex]\[ x = \frac{-1 \pm \sqrt{3}i}{2} \][/tex]
Therefore, the roots are:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{3}i}{2} \quad \text{and} \quad x = -\frac{1}{2} - \frac{\sqrt{3}i}{2} \][/tex]
### b. [tex]\( x^2 + x + 3 = 0 \)[/tex]
Using the quadratic formula for [tex]\( x^2 + x + 3 = 0 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{-11}}{2} \][/tex]
Since [tex]\(\sqrt{-11} = \sqrt{11}i\)[/tex], we get:
[tex]\[ x = \frac{-1 \pm \sqrt{11}i}{2} \][/tex]
Therefore, the roots are:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{11}i}{2} \quad \text{and} \quad x = -\frac{1}{2} - \frac{\sqrt{11}i}{2} \][/tex]
### c. [tex]\( 2x^2 + 7x + 1 = 0 \)[/tex]
Using the quadratic formula for [tex]\( 2x^2 + 7x + 1 = 0 \)[/tex]:
[tex]\[ x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-7 \pm \sqrt{49 - 8}}{4} \][/tex]
[tex]\[ x = \frac{-7 \pm \sqrt{41}}{4} \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-7 + \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-7 - \sqrt{41}}{4} \][/tex]
### d. [tex]\( x^4 - 1 = 0 \)[/tex]
This can be factored as:
[tex]\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1) \][/tex]
Solving each factor separately, we get:
[tex]\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \][/tex]
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
For [tex]\( x^2 + 1 = 0 \)[/tex]:
[tex]\[ x^2 + 1 = 0 \quad \Rightarrow \quad x^2 = -1 \quad \Rightarrow \quad x = \pm i \][/tex]
So, the roots are:
[tex]\[ x = 1, -1, i, -i \][/tex]
### e. [tex]\( x^2 + 9 = 0 \)[/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 9 = 0 \quad \Rightarrow \quad x^2 = -9 \quad \Rightarrow \quad x = \pm \sqrt{-9} \quad \Rightarrow \quad x = \pm 3i \][/tex]
So, the roots are:
[tex]\[ x = 3i \quad \text{and} \quad x = -3i \][/tex]
In summary:
a. [tex]\( x = -\frac{1}{2} + \frac{\sqrt{3}i}{2} \)[/tex] and [tex]\( x = -\frac{1}{2} - \frac{\sqrt{3}i}{2} \)[/tex]
b. [tex]\( x = -\frac{1}{2} + \frac{\sqrt{11}i}{2} \)[/tex] and [tex]\( x = -\frac{1}{2} - \frac{\sqrt{11}i}{2} \)[/tex]
c. [tex]\( x = \frac{-7 + \sqrt{41}}{4} \)[/tex] and [tex]\( x = \frac{-7 - \sqrt{41}}{4} \)[/tex]
d. [tex]\( x = 1, -1, i, -i \)[/tex]
e. [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex]
