2016/116) Find the real roots of the equation

$\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}} = 1$


Solution


let $x-1 = y^2$
so we get  $\sqrt{y^2+4-4y} + \sqrt{y^2+9-6y} = 1$
or $\sqrt{(y-2)^2} + \sqrt{(y-3)^2} = 1$
or $|y-2| + |y-3| = 1$
as 3-2 is 1 so $2 <=y <=3$ or $5  <= x <= 10$