Animated Solution for Mathematics - Quadratic Equations: If the value of real number $a>0$ for which $x^2-5ax+1=0$ and $x^2-ax-5=0$ have a common real roots is $\frac{3}{\sqrt{2\beta }}$, then $\beta$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are exploring the elegant geometry of intersection.

Imagine two parabolas, $y=x^2-5ax+1$ and $y=x^2-ax-5$. They are distinct curves, each carving its own path through the Cartesian plane.

The problem states they share a common real root. This means that at some specific point on the $x$-axis, both parabolas touch the ground at the exact same location. This is our point of entry.

Let us call this mysterious common root $\alpha$. Since $\alpha$ is a root for both equations, it must satisfy them perfectly. We can write this as:

Think of these as two constraints on the same variable. We have a system of two equations, and our goal is to find the value of $a$.

In algebra, when you see a common term that is making your life difficult, the most powerful move is often to eliminate it. By subtracting the second equation from the first, we perform a beautiful act of simplification:

The ${\alpha }^2$ terms vanish into thin air, leaving us with a linear equation: $-4a\alpha +6=0$. Just like that, the complexity collapses. We have isolated the relationship between our root and our parameter:

Now that we have $\alpha$ in terms of $a$, we have built a bridge. We can now return to one of our original equations—let's pick the second one, ${\alpha }^2-a\alpha -5=0$—and substitute our expression for $\alpha$.

We replace $\alpha$ with $\frac{3}{2a}$:

Watch closely as we expand this. The first term becomes $\frac{9}{4a^2}$. The second term is even more satisfying: the $a$ in the numerator and the $a$ in the denominator cancel out perfectly, leaving us with $-\frac{3}{2}$.

Our equation is now:

Combining the constants $-\frac{3}{2}$ and $-5$ gives us $-\frac{13}{2}$. So, we have:

A quick cross-multiplication leads us to $a^2=\frac{9}{26}$. Taking the square root, we get $a=\pm \frac{3}{\sqrt{26}}$.

But remember, the problem gave us the constraint $a>0$. We discard the negative root, leaving us with $a=\frac{3}{\sqrt{26}}$.

Finally, we compare this to the form given in the question, $\frac{3}{\sqrt{2\beta }}$. By inspection, $\sqrt{26}=\sqrt{2\beta }$, which implies $2\beta =26$, and thus $\beta =13$.

This problem is a masterclass in strategy. We didn't need to solve for the roots individually; we used the structure of the equations to our advantage.

Whenever you see 'common root' problems, remember: eliminate the quadratic term, find the linear relationship, and substitute back. You have the tools, you have the logic, and now you have the experience.