Key Concepts and Formulas

• Common Root: If a value $x_0$ is a common root of two equations, then substituting $x_0$ into both equations will satisfy them.
• Elimination Method: Subtracting two equations can eliminate a variable, simplifying the system of equations.
• Solving Quadratic Equations: Techniques for solving quadratic equations include factoring, completing the square, and using the quadratic formula.

Step-by-Step Solution

1. State the Given Equations and Common Root

Explanation: We are given two quadratic equations, and we know they share a common real root. Let's denote the common root as $x_0$ and state the equations. $x^2 - 5ax + 1 = 0 \quad \cdots (1)$ $x^2 - ax - 5 = 0 \quad \cdots (2)$ Since $x_0$ is a common root, it must satisfy both equations: $x_0^2 - 5ax_0 + 1 = 0 \quad \cdots (3)$ $x_0^2 - ax_0 - 5 = 0 \quad \cdots (4)$

2. Eliminate the $x_0^2$ Term

Explanation: Subtracting equation (4) from equation (3) will eliminate the $x_0^2$ term, resulting in a linear equation in terms of $x_0$. This allows us to express $x_0$ in terms of $a$. Subtract equation (4) from equation (3): $(x_0^2 - 5ax_0 + 1) - (x_0^2 - ax_0 - 5) = 0$ $x_0^2 - 5ax_0 + 1 - x_0^2 + ax_0 + 5 = 0$ $-4ax_0 + 6 = 0$ $4ax_0 = 6$ $x_0 = \frac{6}{4a} = \frac{3}{2a} \quad \cdots (5)$

3. Substitute the Common Root into Equation (2)

Explanation: Substituting the expression for $x_0$ from equation (5) into either equation (3) or (4) will eliminate $x_0$, leaving an equation solely in terms of $a$. We choose equation (2) here. Substitute $x_0 = \frac{3}{2a}$ into equation (4): $\left(\frac{3}{2a}\right)^2 - a\left(\frac{3}{2a}\right) - 5 = 0$ $\frac{9}{4a^2} - \frac{3}{2} - 5 = 0$

4. Solve for $a^2$

Explanation: Now we solve the above equation for $a^2$.

$\frac{9}{4a^2} = \frac{3}{2} + 5$ $\frac{9}{4a^2} = \frac{3}{2} + \frac{10}{2}$ $\frac{9}{4a^2} = \frac{13}{2}$ $18 = 52a^2$ $a^2 = \frac{18}{52} = \frac{9}{26}$

5. Solve for $a$

Explanation: Since $a>0$, we take the positive square root. $a = \sqrt{\frac{9}{26}} = \frac{3}{\sqrt{26}}$

6. Determine the Value of $\beta$

Explanation: The problem states that $a = \frac{3}{\sqrt{2\beta}}$. We compare this with our value for $a$ to find $\beta$. We have $a = \frac{3}{\sqrt{26}}$ and $a = \frac{3}{\sqrt{2\beta}}$. Therefore, $\frac{3}{\sqrt{2\beta}} = \frac{3}{\sqrt{26}}$ $\sqrt{2\beta} = \sqrt{26}$ $2\beta = 26$ $\beta = \frac{26}{2} = 13$

Common Mistakes & Tips

• Remember to consider the condition $a > 0$ when taking the square root.
• Be careful with algebraic manipulations, especially when dealing with fractions and square roots.
• An alternative method is to eliminate the constant term. Multiply equation (1) by 5 and subtract equation (2).

Summary

We found the common root $x_0$ in terms of $a$ by eliminating the $x_0^2$ term. Then, we substituted this expression back into one of the original equations to obtain an equation in terms of $a$ only. Solving for $a$ and using the given format $a = \frac{3}{\sqrt{2\beta}}$, we determined the value of $\beta$.

The final answer is \boxed{13}.