Rooty Equation

$\sqrt {x + 1} - \sqrt {x - 1} = \sqrt {4x - 1}$

First square both sides

$(x + 1) + (x - 1) \pm 2\sqrt {(-x)^2 + 1} = 4x - 1$

The square root left can be found by re-arranging it

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

Now we have a problem, there's a $\sqrt {-1}$ , when there's one of these there isn't a real answer to the equation so there's no solution

squaring both sides,then again square both sides .answer is 1.25

We can just first square the LHS and RHS of the given equation that way we have just one terms with square root remaining which we can eliminate the square root by arranging the terms on both the sides of the equation and squaring again. After that we just solve the remaining part of the equation and get the answer i.e. no solution. Many people including me had got 5/4 as an solution but we also have to check if it satisfies the given equation or not. On doing that we get that 5/4 does not satisfy the given equation. Hence we get no solutions as the answer.

Just to give people an understanding of why 5/4 is not the answer to this problem, take another example into consideration:

(1/2)x = -√(x-1)

Square both sides and solve, you'll get x = 2. However, look at what you have in the original equation in terms of their graphs. On the left, you have a diagonal line. On the right, you have the lower half of a sideways parabola. When you square both sides though, you have effectively introduced the upper half of the sideways parabola into the equation. If you plug in 2 to the line immediately after squaring both sides, things work out. But plug it into the original, you get 1 = -1.

Something similar is happening in this problem. Algebraically things might seem to work out, but you must take the domain and range of both sides of the original problem into consideration to get a proper result.