by setting y = x + x^-1, find the solutions of 
x^4+10x^3 +26x^2+10x + 1 = 0.

To solve the quartic equation \(x^4 + 10x^3 + 26x^2 + 10x + 1 = 0\) using the substitution \(y = x + \frac{1}{x}\), we first need to express the given quartic in terms of \(y\). 

### Step 1: Express \(x + \frac{1}{x}\) and related terms.
We can manipulate the substitution to help simplify the quartic equation.

#### Step 2: Calculate \(x^2 + \frac{1}{x^2}\) in terms of \(y\).
We know:
\[
y = x + \frac{1}{x}.
\]
Now square both sides:
\[
y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2},
\]
so
\[
x^2 + \frac{1}{x^2} = y^2 - 2.
\]

#### Step 3: Calculate \(x^3 + \frac{1}{x^3}\) in terms of \(y\).
Now we cube \(x + \frac{1}{x}\):
\[
y^3 = \left(x + \frac{1}{x}\right)^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3},
\]
which simplifies to:
\[
y^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = x^3 + \frac{1}{x^3} + 3y.
\]
Thus,
\[
x^3 + \frac{1}{x^3} = y^3 - 3y.
\]

#### Step 4: Rewrite the equation in terms of \(y\).
The given equation is:
\[
x^4 + 10x^3 + 26x^2 + 10x + 1 = 0.
\]
Multiply both sides by \(\frac{1}{x^2}\) to get:
\[
\left(x^2 + \frac{1}{x^2}\right) + 10\left(x + \frac{1}{x}\right) + 26 = 0.
\]
Substitute \(x + \frac{1}{x} = y\) and \(x^2 + \frac{1}{x^2} = y^2 - 2\):
\[
(y^2 - 2) + 10y + 26 = 0.
\]
Simplify this to:
\[
y^2 + 10y + 24 = 0.
\]

#### Step 5: Solve the quadratic equation for \(y\).
Solve the quadratic equation \(y^2 + 10y + 24 = 0\) using the quadratic formula:
\[
y = \frac{-10 \pm \sqrt{10^2 - 4(1)(24)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 96}}{2} = \frac{-10 \pm \sqrt{4}}{2} = \frac{-10 \pm 2}{2}.
\]
So the two solutions for \(y\) are:
\[
y = \frac{-10 + 2}{2} = -4 \quad \text{and} \quad y = \frac{-10 - 2}{2} = -6.
\]

#### Step 6: Find \(x\) from \(y = x + \frac{1}{x}\).
Now we solve \(x + \frac{1}{x} = -4\) and \(x + \frac{1}{x} = -6\).

For \(x + \frac{1}{x} = -4\), multiply both sides by \(x\):
\[
x^2 + 1 = -4x \quad \Rightarrow \quad x^2 + 4x + 1 = 0.
\]
The solutions to this quadratic equation are:
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2}.
\]
Thus, \(x = -2 + \sqrt{3}\) and \(x = -2 - \sqrt{3}\).

For \(x + \frac{1}{x} = -6\), multiply both sides by \(x\):
\[
x^2 + 1 = -6x \quad \Rightarrow \quad x^2 + 6x + 1 = 0.
\]
The solutions to this quadratic equation are:
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(1)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2}.
\]
Thus, \(x = -3 + 2\sqrt{2}\) and \(x = -3 - 2\sqrt{2}\).

### Final Answer:
The solutions to the equation \(x^4 + 10x^3 + 26x^2 + 10x + 1 = 0\) are:
\[
x = -2 + \sqrt{3}, \quad x = -2 - \sqrt{3}, \quad x = -3 + 2\sqrt{2}, \quad x = -3 - 2\sqrt{2}.
\]

Would you like further details on any step?

Here are 5 related questions:
1. How does the substitution \(y = x + \frac{1}{x}\) simplify quartic equations?
2. Can you derive the cubic identity used to find \(x^3 + \frac{1}{x^3}\)?
3. How does the discriminant help in determining the nature of solutions?
4. What are other methods to solve quartic equations without substitution?
5. How can these techniques be applied to polynomials of higher degree?

**Tip**: When solving equations involving substitutions, always remember to express all terms of the equation in terms of the new variable before solving.

By setting y = x + x^-1, find the solutions of x^4 + 10x^3 + 26x^2 + 10x + 1 = 0.

Learn how to solve the quartic equation x^4 + 10x^3 + 26x^2 + 10x + 1 = 0 using the substitution y = x + 1/x. This detailed solution covers key algebraic concepts such as substitution in polynomial equations and the quadratic formula. Ideal for students in Grades 10-12.

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