Answer :
Sure! Let's solve both equations step-by-step:
9. Equation: [tex]\(x^4 - 9x^2 + 8 = 0\)[/tex]
1. Notice that this equation can be set up as a quadratic in form by substituting [tex]\(y = x^2\)[/tex]. The equation becomes:
[tex]\[
y^2 - 9y + 8 = 0
\][/tex]
2. Now, let's factor this quadratic:
[tex]\[
(y - 8)(y - 1) = 0
\][/tex]
3. This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[
y = 8 \quad \text{and} \quad y = 1
\][/tex]
4. Substitute back [tex]\(y = x^2\)[/tex]:
- For [tex]\(y = 8\)[/tex], solve [tex]\(x^2 = 8\)[/tex]:
[tex]\[
x = \sqrt{8} \quad \text{or} \quad x = -\sqrt{8}
\][/tex]
Simplifying, we get:
[tex]\[
x = 2\sqrt{2} \quad \text{or} \quad x = -2\sqrt{2}
\][/tex]
- For [tex]\(y = 1\)[/tex], solve [tex]\(x^2 = 1\)[/tex]:
[tex]\[
x = 1 \quad \text{or} \quad x = -1
\][/tex]
5. So the roots are:
[tex]\[
x = 2\sqrt{2}, \, -2\sqrt{2}, \, 1, \, -1
\][/tex]
10. Equation: [tex]\(x^4 + 7x^3 + 14x^2 + 6x = 0\)[/tex]
1. Notice that [tex]\(x\)[/tex] can be factored out from every term:
[tex]\[
x(x^3 + 7x^2 + 14x + 6) = 0
\][/tex]
2. This gives the first root:
[tex]\[
x = 0
\][/tex]
3. Now solve the cubic equation: [tex]\(x^3 + 7x^2 + 14x + 6 = 0\)[/tex]
4. We can look for rational roots using the Rational Root Theorem; test possible factors of 6. However, trial and error or use of a tool like Desmos can help find one root. Let's say we found:
- [tex]\(x = -1\)[/tex] is a root.
5. Use synthetic division or polynomial division to divide the cubic by [tex]\(x + 1\)[/tex], yielding:
[tex]\[
x^3 + 7x^2 + 14x + 6 = (x + 1)(x^2 + 6x + 6)
\][/tex]
6. Solve [tex]\(x^2 + 6x + 6 = 0\)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 6}}{2 \times 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 - 24}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{12}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2\sqrt{3}}{2}
\][/tex]
[tex]\[
x = -3 \pm \sqrt{3}
\][/tex]
7. Therefore, the roots of the second equation are:
[tex]\[
x = 0, \, -1, \, -3 + \sqrt{3}, \, -3 - \sqrt{3}
\][/tex]
These are the roots for both equations. If anything is confusing, feel free to ask for further clarification!
9. Equation: [tex]\(x^4 - 9x^2 + 8 = 0\)[/tex]
1. Notice that this equation can be set up as a quadratic in form by substituting [tex]\(y = x^2\)[/tex]. The equation becomes:
[tex]\[
y^2 - 9y + 8 = 0
\][/tex]
2. Now, let's factor this quadratic:
[tex]\[
(y - 8)(y - 1) = 0
\][/tex]
3. This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[
y = 8 \quad \text{and} \quad y = 1
\][/tex]
4. Substitute back [tex]\(y = x^2\)[/tex]:
- For [tex]\(y = 8\)[/tex], solve [tex]\(x^2 = 8\)[/tex]:
[tex]\[
x = \sqrt{8} \quad \text{or} \quad x = -\sqrt{8}
\][/tex]
Simplifying, we get:
[tex]\[
x = 2\sqrt{2} \quad \text{or} \quad x = -2\sqrt{2}
\][/tex]
- For [tex]\(y = 1\)[/tex], solve [tex]\(x^2 = 1\)[/tex]:
[tex]\[
x = 1 \quad \text{or} \quad x = -1
\][/tex]
5. So the roots are:
[tex]\[
x = 2\sqrt{2}, \, -2\sqrt{2}, \, 1, \, -1
\][/tex]
10. Equation: [tex]\(x^4 + 7x^3 + 14x^2 + 6x = 0\)[/tex]
1. Notice that [tex]\(x\)[/tex] can be factored out from every term:
[tex]\[
x(x^3 + 7x^2 + 14x + 6) = 0
\][/tex]
2. This gives the first root:
[tex]\[
x = 0
\][/tex]
3. Now solve the cubic equation: [tex]\(x^3 + 7x^2 + 14x + 6 = 0\)[/tex]
4. We can look for rational roots using the Rational Root Theorem; test possible factors of 6. However, trial and error or use of a tool like Desmos can help find one root. Let's say we found:
- [tex]\(x = -1\)[/tex] is a root.
5. Use synthetic division or polynomial division to divide the cubic by [tex]\(x + 1\)[/tex], yielding:
[tex]\[
x^3 + 7x^2 + 14x + 6 = (x + 1)(x^2 + 6x + 6)
\][/tex]
6. Solve [tex]\(x^2 + 6x + 6 = 0\)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 6}}{2 \times 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 - 24}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{12}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2\sqrt{3}}{2}
\][/tex]
[tex]\[
x = -3 \pm \sqrt{3}
\][/tex]
7. Therefore, the roots of the second equation are:
[tex]\[
x = 0, \, -1, \, -3 + \sqrt{3}, \, -3 - \sqrt{3}
\][/tex]
These are the roots for both equations. If anything is confusing, feel free to ask for further clarification!
