Animated Solution for Mathematics - Quadratic Equations: If $\alpha$ and $\beta (\alpha <\beta )$ are the roots of the equation $(-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6\sqrt{x})+(9-2\sqrt{3})=0,x\ge 0$, then $\sqrt{\frac{\beta }{\alpha }}+\sqrt{\alpha \beta }$ is equal to :

Imagine you are standing before a complex, intimidating equation:

At first glance, it looks like a chaotic mess of square roots and absolute values. But in the world of JEE Advanced, chaos is often just order in disguise. Our goal today is to peel back the layers of this expression to reveal the elegant quadratic structure hiding beneath.

Whenever you see $\sqrt{x}$ appearing repeatedly, your first instinct should be to simplify the landscape. Let us define a new variable, $t=\sqrt{x}$.

Since $x\ge 0$, we know that $t$ must be non-negative ($t\ge 0$). Our equation now transforms into:

Look closely at the terms $(t^2-6t+9)$. This is the expansion of $(t-3{)}^2$. We have a perfect square sitting right there, waiting to be recognized.

We know that for any real number $y$, $y^2=|y{|}^2$. Therefore, we can rewrite $(t-3{)}^2$ as $|t-3{|}^2$.

Now, watch how the equation cleans up when we group the terms:

By letting $u=|t-3|$, we have successfully reduced a terrifying radical equation into a simple, beautiful quadratic:

Now, we factorize. We need two numbers that multiply to $-2\sqrt{3}$ and add to $\sqrt{3}-2$. Those numbers are $\sqrt{3}$ and $-2$.

Thus, the equation becomes:

This gives us two potential paths: $u=2$ or $u=-\sqrt{3}$. However, we must remember our constraint: $u=|t-3|$, which must be $\ge 0$.

We must reject $u=-\sqrt{3}$ because an absolute value cannot be negative. We are left with the only valid solution: $u=2$.

Now we backtrack. Since $u=|t-3|=2$, we have two cases for $t$:

1. $t-3=2⟹t=5$ 2. $t-3=-2⟹t=1$

Since $t=\sqrt{x}$, we square these values to find $x$: $x=25$ and $x=1$. Given the condition $\alpha <\beta$, we identify $\alpha =1$ and $\beta =25$.

Finally, we calculate the expression $\sqrt{\frac{\beta }{\alpha }}+\sqrt{\alpha \beta }$:

And there it is—the final answer is 10. What started as a daunting expression collapsed into a simple, elegant result.