Question

$\sqrt{y+1} + \sqrt{y-1} =\sqrt{4y-1}$ find y

Answer 1

Answer:

$\huge\boxed{\boxed{\red{Solution:-}}}$

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

squaring both sides

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

$\bf\implies{2y} - 2\sqrt{(y^2 - 1)}= {4y - 1}$

$\bf\implies{2}\sqrt{y^2 - 1} = {4y - 2y -1}$

$\bf\implies{2}\sqrt{y^2-1} = {2y - 1}$

Again squaring both sides.

$\bf\implies{4(y^2 -1 ) = 4y^2 - 1 - 4y}$

$\bf\implies{4y^2 - 4 = 4y^2 - 1 - 4y}$

$\bf\implies{4y = 5}$

$\bf\implies{y} =\frac{5}{4}$

$\bf{\red{\underline{\underline{MuskaŊ}}}}$