Understanding the Problem

The question asks us to find the value of a specific algebraic expression given a quadratic equation. The equation is $x^2 - 3x + 1 = 0$, and the expression is $\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}$. We need to use the given equation to simplify the expression and find its numerical value.

Key Relationship from the Equation

The given equation is $x^2 - 3x + 1 = 0$. This equation relates $x$, $x^2$, and the constant term. A useful technique when dealing with expressions involving $x$ and $\frac{1}{x}$ is to divide the equation by $x$. Since $x=0$ would give $0-0+1=1 \neq 0$, $x$ cannot be zero, so we can safely divide by $x$.

Dividing the equation $x^2 - 3x + 1 = 0$ by $x$, we get:

$\frac{x^2}{x} - \frac{3x}{x} + \frac{1}{x} = \frac{0}{x}$

$x - 3 + \frac{1}{x} = 0$

Rearranging this, we find a key relationship:

$x + \frac{1}{x} = 3$

This relationship is very useful for simplifying expressions involving powers of $x$ and $\frac{1}{x}$.

Simplifying the Denominator

The denominator of the given expression is $x^2 + 5x + 1$. We can use the original equation $x^2 - 3x + 1 = 0$ to simplify this.

From $x^2 - 3x + 1 = 0$, we can rearrange it to find the value of $x^2 + 1$:

$x^2 + 1 = 3x$

Now, substitute this into the denominator:

Denominator $= (x^2 + 1) + 5x$

Substitute $x^2 + 1 = 3x$:

Denominator $= 3x + 5x = 8x$

So, the denominator simplifies to $8x$.

Simplifying the Expression using the Denominator

The original expression is $\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}$. We found the denominator simplifies to $8x$. So the expression becomes:

$\frac{x^4 + \frac{1}{x^2}}{8x}$

We can split this fraction into two terms:

$\frac{x^4}{8x} + \frac{\frac{1}{x^2}}{8x}$

Simplify each term:

• $\frac{x^4}{8x} = \frac{x^{4-1}}{8} = \frac{x^3}{8}$
• $\frac{\frac{1}{x^2}}{8x} = \frac{1}{x^2 \cdot 8x} = \frac{1}{8x^{2+1}} = \frac{1}{8x^3}$

So the expression is equal to:

$\frac{x^3}{8} + \frac{1}{8x^3} = \frac{1}{8} \left( x^3 + \frac{1}{x^3} \right)$

Now, the problem reduces to finding the value of $x^3 + \frac{1}{x^3}$.

Calculating $x^3 + \frac{1}{x^3}$

We know that $x + \frac{1}{x} = 3$. We can use the algebraic identity for the sum of cubes: $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.

Let $a = x$ and $b = \frac{1}{x}$. Then $ab = x \cdot \frac{1}{x} = 1$.

Using the identity:

$x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x \cdot \frac{1}{x} \right) \left( x + \frac{1}{x} \right)$

$x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right)$

Substitute the value $x + \frac{1}{x} = 3$ into this equation:

$x^3 + \frac{1}{x^3} = (3)^3 - 3(3)$

$x^3 + \frac{1}{x^3} = 27 - 9$

$x^3 + \frac{1}{x^3} = 18$

Calculating the Final Value of the Expression

We found that the expression is equal to $\frac{1}{8} \left( x^3 + \frac{1}{x^3} \right)$.

Substitute the value $x^3 + \frac{1}{x^3} = 18$ into this:

Expression Value $= \frac{1}{8} (18)$

Expression Value $= \frac{18}{8}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

Expression Value $= \frac{18 \div 2}{8 \div 2} = \frac{9}{4}$

Final Answer

The value of the expression $\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}$ is $\frac{9}{4}$.

Revision Table: Key Algebraic Identities

Understanding algebraic identities is crucial for solving such problems efficiently. Here are some relevant identities:

Additional Information: Generalizing the Method

The technique used here can be generalized for other powers. If you have $x + \frac{1}{x} = k$, you can find expressions for $x^n + \frac{1}{x^n}$ or $x^n - \frac{1}{x^n}$ for various integer values of $n$.

• For $n=2$: $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = k^2 - 2$.
• For $n=3$: $x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = k^3 - 3k$.
• For $n=4$: $x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (k^2 - 2)^2 - 2$.

This method provides a powerful way to solve problems involving powers of $x$ and $\frac{1}{x}$ when a linear relationship like $x \pm \frac{1}{x} = k$ is derived from a given equation.