Animated Solution for Mathematics - Quadratic Equations: Solve for $x:\sqrt{x+1}-\sqrt{x-1}=1$.

A Journey Through Radicals

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving an equation; we are embarking on a quest to uncover the hidden value of $x$ in the expression $\sqrt{x+1}-\sqrt{x-1}=1$.

This might look like a simple problem, but it is a classic trap for the unwary. Let us navigate this together.

Defining the Domain

Before we even touch our pens to paper, we must respect the laws of the land. We are dealing with square roots, and in the realm of real numbers, the expression inside a square root must be non-negative.

For the first term, $\sqrt{x+1}$, we require $x+1\ge 0$, which simplifies to $x\ge -1$. For the second term, $\sqrt{x-1}$, we require $x-1\ge 0$, which simplifies to $x\ge 1$.

To satisfy both conditions simultaneously, our combined domain is $x\ge 1$. This is our boundary of existence; any solution we find must live within this region.

Now, how do we dismantle this radical structure? A common mistake is to square both sides immediately. If you do that, you will find yourself wrestling with a messy cross-term: $-2\sqrt{(x+1)(x-1)}$.

Instead, let us use a more elegant strategy: isolation. By rearranging the equation to $\sqrt{x+1}=1+\sqrt{x-1}$, we position ourselves for a much cleaner algebraic dance.

We have moved the negative radical to the right, setting the stage for a successful squaring operation.

Now, we square both sides: $(\sqrt{x+1}{)}^2=(1+\sqrt{x-1}{)}^2$. On the left, the square and the root vanish, leaving us with $x+1$.

On the right, we invoke the powerful identity $(a+b{)}^2=a^2+b^2+2ab$. Here, $a=1$ and $b=\sqrt{x-1}$. Expanding this, we get $1^2+(\sqrt{x-1}{)}^2+2(1)(\sqrt{x-1})$, which simplifies to $1+(x-1)+2\sqrt{x-1}$.

Look at that! The $1$ and $-1$ cancel out, leaving us with $x+1=x+2\sqrt{x-1}$. The variable $x$ appears on both sides, and with a simple subtraction, it vanishes entirely. We are left with the beautifully simple:

We are almost there. We have one radical remaining, so we repeat our strategy: square both sides again. This gives us $1^2=(2\sqrt{x-1}{)}^2$, which simplifies to $1=4(x-1)$.

Distributing the $4$, we get $1=4x-4$. Adding $4$ to both sides yields $5=4x$, and finally, dividing by $4$ gives us:

In the world of JEE, verification is not optional; it is the final seal of truth. Let us plug $x=1.25$ back into our original equation:

The left-hand side perfectly matches the right-hand side. We have conquered the radical, respected the domain, and verified our path. The final answer is $x=1.25$.