Answer :
Certainly! Let's solve each of these equations step by step.
### a. [tex]\(x^2 + x + 1 = 0\)[/tex]
To solve the quadratic equation [tex]\(x^2 + x + 1 = 0\)[/tex], we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \][/tex]
2. Since the discriminant is negative, the roots are complex:
[tex]\[ x = \frac{-1 \pm \sqrt{-3}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{3}}{2} \][/tex]
3. So the roots are:
[tex]\[ x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \quad \text{and} \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
### d. [tex]\(x^2 + x + 3 = 0\)[/tex]
We use the quadratic formula for [tex]\(x^2 + x + 3 = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = 3\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 3 = 1 - 12 = -11 \][/tex]
2. Since the discriminant is negative, the roots are complex:
[tex]\[ x = \frac{-1 \pm \sqrt{-11}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{11}}{2} \][/tex]
3. So the roots are:
[tex]\[ x = -\frac{1}{2} - \frac{\sqrt{11}}{2}i \quad \text{and} \quad x = -\frac{1}{2} + \frac{\sqrt{11}}{2}i \][/tex]
### 3. [tex]\(2x^2 + 7x + 1 = 0\)[/tex]
We use the quadratic formula for [tex]\(2x^2 + 7x + 1 = 0\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 1 = 49 - 8 = 41 \][/tex]
2. Since the discriminant is positive, the roots are real:
[tex]\[ x = \frac{-7 \pm \sqrt{41}}{2 \cdot 2} = \frac{-7 \pm \sqrt{41}}{4} \][/tex]
3. So the roots are:
[tex]\[ x = \frac{-7 - \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-7 + \sqrt{41}}{4} \][/tex]
### e. [tex]\(x^4 - 1 = 0\)[/tex]
This can be factored as:
[tex]\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \][/tex]
1. Factor further:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
[tex]\[ x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm i \][/tex]
2. So the roots are:
[tex]\[ x = 1, -1, i, \text{and} -i \][/tex]
### c. [tex]\(x^2 + 9 = 0\)[/tex]
To find the roots, solve [tex]\(x^2 + 9 = 0\)[/tex].
1. Rewrite the equation:
[tex]\[ x^2 = -9 \][/tex]
2. Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{-9} = \pm 3i \][/tex]
3. So the roots are:
[tex]\[ x = 3i \quad \text{and} \quad x = -3i \][/tex]
### Summary of Roots:
1. [tex]\(x^2 + x + 1 = 0\)[/tex] ⟹ [tex]\(x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex] and [tex]\(x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex]
2. [tex]\(x^2 + x + 3 = 0\)[/tex] ⟹ [tex]\(x = -\frac{1}{2} - \frac{\sqrt{11}}{2}i\)[/tex] and [tex]\(x = -\frac{1}{2} + \frac{\sqrt{11}}{2}i\)[/tex]
3. [tex]\(2x^2 + 7x + 1 = 0\)[/tex] ⟹ [tex]\(x = \frac{-7 - \sqrt{41}}{4}\)[/tex] and [tex]\(x = \frac{-7 + \sqrt{41}}{4}\)[/tex]
4. [tex]\(x^4 - 1 = 0\)[/tex] ⟹ [tex]\(x = 1, -1, i, -i\)[/tex]
5. [tex]\(x^2 + 9 = 0\)[/tex] ⟹ [tex]\(x = 3i\)[/tex] and [tex]\(x = -3i\)[/tex]
### a. [tex]\(x^2 + x + 1 = 0\)[/tex]
To solve the quadratic equation [tex]\(x^2 + x + 1 = 0\)[/tex], we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \][/tex]
2. Since the discriminant is negative, the roots are complex:
[tex]\[ x = \frac{-1 \pm \sqrt{-3}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{3}}{2} \][/tex]
3. So the roots are:
[tex]\[ x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \quad \text{and} \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
### d. [tex]\(x^2 + x + 3 = 0\)[/tex]
We use the quadratic formula for [tex]\(x^2 + x + 3 = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = 3\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 3 = 1 - 12 = -11 \][/tex]
2. Since the discriminant is negative, the roots are complex:
[tex]\[ x = \frac{-1 \pm \sqrt{-11}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{11}}{2} \][/tex]
3. So the roots are:
[tex]\[ x = -\frac{1}{2} - \frac{\sqrt{11}}{2}i \quad \text{and} \quad x = -\frac{1}{2} + \frac{\sqrt{11}}{2}i \][/tex]
### 3. [tex]\(2x^2 + 7x + 1 = 0\)[/tex]
We use the quadratic formula for [tex]\(2x^2 + 7x + 1 = 0\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 1 = 49 - 8 = 41 \][/tex]
2. Since the discriminant is positive, the roots are real:
[tex]\[ x = \frac{-7 \pm \sqrt{41}}{2 \cdot 2} = \frac{-7 \pm \sqrt{41}}{4} \][/tex]
3. So the roots are:
[tex]\[ x = \frac{-7 - \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-7 + \sqrt{41}}{4} \][/tex]
### e. [tex]\(x^4 - 1 = 0\)[/tex]
This can be factored as:
[tex]\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \][/tex]
1. Factor further:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
[tex]\[ x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm i \][/tex]
2. So the roots are:
[tex]\[ x = 1, -1, i, \text{and} -i \][/tex]
### c. [tex]\(x^2 + 9 = 0\)[/tex]
To find the roots, solve [tex]\(x^2 + 9 = 0\)[/tex].
1. Rewrite the equation:
[tex]\[ x^2 = -9 \][/tex]
2. Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{-9} = \pm 3i \][/tex]
3. So the roots are:
[tex]\[ x = 3i \quad \text{and} \quad x = -3i \][/tex]
### Summary of Roots:
1. [tex]\(x^2 + x + 1 = 0\)[/tex] ⟹ [tex]\(x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex] and [tex]\(x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex]
2. [tex]\(x^2 + x + 3 = 0\)[/tex] ⟹ [tex]\(x = -\frac{1}{2} - \frac{\sqrt{11}}{2}i\)[/tex] and [tex]\(x = -\frac{1}{2} + \frac{\sqrt{11}}{2}i\)[/tex]
3. [tex]\(2x^2 + 7x + 1 = 0\)[/tex] ⟹ [tex]\(x = \frac{-7 - \sqrt{41}}{4}\)[/tex] and [tex]\(x = \frac{-7 + \sqrt{41}}{4}\)[/tex]
4. [tex]\(x^4 - 1 = 0\)[/tex] ⟹ [tex]\(x = 1, -1, i, -i\)[/tex]
5. [tex]\(x^2 + 9 = 0\)[/tex] ⟹ [tex]\(x = 3i\)[/tex] and [tex]\(x = -3i\)[/tex]
