$\sqrt{y + 1} - \sqrt{y - 1} = \sqrt{4y - 1}$. Find the value of $y$.

Question

I tried this way Squaring both sides

$(y+1)+ (y-1) - 2\sqrt{(y+1)(y-1)}=(4y-1)$

$1-2y =2\sqrt{(y+1)(y-1)}$

Again squaring both side.

$1+4y^2-4y =4(y^2-1)$
$4y^2-4y+1=4y^2-4$
$y=5/4$

But on putting the value result was.
1=2.
Which was wrong.
I want to know where is the mistake.

Answer 1

Consider function on LHS, $\sqrt{y+1} - \sqrt{y-1}$. It is not defined for $y<1$ and has a maximum value at $y=1$ as it is a decreasing function beyond $y\ge1$. Hence, its maximum value is $\sqrt{2}$.

On the other hand, RHS $\sqrt{4y-1}$ is an increasing function and is defined only for $y\ge \frac{1}{4}$. At $y=1$, RHS is $\sqrt3$, which is greater than the maximum value of LHS.

Hence, LHS and RHS never meet at any point. The equation has no solution. See the graph for clarity.

Coming back to your question, it is important to note that when you square an equation, it can introduce additional roots. Consequently, the presence of this extra root in the resulting equation is what you are observing, which was not present in the original equation.

Answer 2

Here we have

$\displaystyle \sqrt{y+1}-\sqrt{y-1}\leq \sqrt{y+1}<\sqrt{4y-1}$

Which is true when $\displaystyle y>\frac{2}{3}$

Above equality hold when $y=1$

So equation has no solution

Answer 3

Another way, for $y=1$ equality doesn't hold and for $y>1$

$$\sqrt{y + 1} - \sqrt{y - 1} = \sqrt{4y - 1} \iff \frac2{\sqrt{y + 1} + \sqrt{y - 1}}=\sqrt{4y - 1} $$

$$\iff \sqrt{(4y - 1)(y+1)}+\sqrt{(4y - 1)(y-1)}=2$$

but

$$f(y)=\sqrt{(4y - 1)(y+1)}+\sqrt{(4y - 1)(y-1)} \ge f(1)=\sqrt 6>2$$