Animated Solution for Mathematics - Basic Mathematics: Show that the square of $\frac{\sqrt{26-15\sqrt{3}}}{5\sqrt{2}-\sqrt{38+5\sqrt{3}}}$ is a rational number.

Welcome, future engineer. Today, we stand before a problem that, at first glance, looks like a tangled mess of roots and irrational numbers. You see the expression:

Your instinct might be to panic, but in the world of JEE Advanced, complexity is often just a mask for hidden simplicity. Our goal is to prove that the square of this expression is a rational number.

The Numerator and the Golden Trick

Let us isolate the numerator: $\sqrt{26-15\sqrt{3}}$. The problem here is that $15\sqrt{3}$ does not have an even coefficient.

To use the identity $(a-b{)}^2=a^2+b^2-2ab$, we desperately need that $2$. We perform the 'Golden Trick' of nested radicals by multiplying and dividing by $2$:

Now, look at the numerator $52-30\sqrt{3}$. We need to find $a$ and $b$ such that $2ab=30\sqrt{3}$.

If we set $a=3\sqrt{3}$ and $b=5$, let us check the sum of their squares: $(3\sqrt{3}{)}^2+5^2=27+25=52$. It is a perfect match! Thus, the numerator simplifies to:

The Denominator and the Pattern

Now, we turn to the denominator's nested radical: $\sqrt{38+5\sqrt{3}}$. We apply the same logic, multiplying and dividing by $2$ to obtain:

We need $2ab=10\sqrt{3}$, so $ab=5\sqrt{3}$. Let us test $a=5\sqrt{3}$ and $b=1$.

The sum of squares is $(5\sqrt{3}{)}^2+1^2=75+1=76$. Again, it fits perfectly! The denominator radical becomes:

The Assembly

Now, we substitute these simplified forms back into our original expression $x$:

We can clear the double-decker fraction by multiplying the numerator and denominator by $\sqrt{2}$:

Simplifying the denominator, we get $10-5\sqrt{3}-1$, which is $9-5\sqrt{3}$. So, our expression is:

The Grand Finale

The final step is to square $x$. We have:

Expanding the numerator gives $27+25-30\sqrt{3}=52-30\sqrt{3}$. Expanding the denominator gives $81+75-90\sqrt{3}=156-90\sqrt{3}$.

Now, look closely at the denominator: $156-90\sqrt{3}$. If we factor out a $3$, we get $3(52-30\sqrt{3})$.

The numerator and the denominator share the exact same irrational term! They cancel out, leaving us with:

We have arrived. The result is a rational number. This problem teaches us that no matter how intimidating the expression, if you follow the logical structure of algebra, the chaos will always resolve into order.