x can also be presented as a + 1/2 - (a-3/4)^1/2. And dx/da=1 - 1/[2*(a-3/4)^1/2], which is positive when a>1. Therefore x has a minimum value of 1, when a=1.
If a = 1, then x = 1 is clearly a solution. Suppose a >1. Let F(x) := x + sqrt(x-1) - a, for x >= 1. F(x) is clearly a real, continuous function. F(1) = 1 -a < 0. F(a) = sqrt(a-1) > 0. So, F has at least one zero. The video shows that x = (2a +1 - sqrt(4a - 3))/2 is the only possible zero of F and therefore it must be a zero of F.
x can also be presented as a + 1/2 - (a-3/4)^1/2.
And dx/da=1 - 1/[2*(a-3/4)^1/2], which is positive when a>1. Therefore x has a minimum value of 1, when a=1.
x+√(x-1)=a
In this equation x-1≥0 because of the square root; x≥1. x and √(x-1) are the increasing functions, so x+√(x-1)≥1+0; a≥1
If a = 1, then x = 1 is clearly a solution. Suppose a >1. Let
F(x) := x + sqrt(x-1) - a, for x >= 1. F(x) is clearly a real, continuous
function. F(1) = 1 -a < 0. F(a) = sqrt(a-1) > 0. So, F has at least one zero. The video shows that x = (2a +1 - sqrt(4a - 3))/2 is the only possible zero of F and therefore it must be a zero of F.