Root(y+1) -root(y-1)=root(4y-1) find y

Question

Theory of Quadratic equations has to be used.

Answer 1

√(y+1)-√(y-1)=√(4y-1).

Square both sides: y+1-2√(y^2-1)+y-1=4y-1.

-2√(y^2-1)=2y-1.

Square again: 4(y^2-1)=4y^2-4y+1.

y^2 terms cancel leaving linear equation: 4y=5 so y=5/4=1.25.

However, if y=5/4 in the original equation: √(9/4)-√(1/4)=3/2-1/2=1≠√(5-1)=2 unless we take the negative square root of y-1 so that 3/2+1/2=2. Yet the 4y^2 term cancels out during the solution workings leaving us with a linear equation which only has one solution. If only the positive square root is implied, there is no solution. Perhaps the question was misstated? Perhaps it should have been √(y+1)+√(y-1)=√(4y-1), which allows the solution y=5/4.