Determine if `sqrt(x)+sqrt(x+9)=2` has any solutions:
`x>=0` because the domain of `sqrt(x)` is nonnegative x.
`sqrt(x+9)>=3` for `x>=0` .
So you have a nonnegative number added to a number greater than or equal to 3 -- their sum cannot be 2. (The sum of the left hand side is greater than or equal to 3.)
There are no roots.
You need to try eliminate the radicals, hence, you need to square both sides, such that:
`(sqrt x + sqrt(x + 9))^2 = 2^2`
Expanding the square, yields:
`x + 2sqrt(x(x+9)) + x + 9 = 4`
You need to isolate the radical to the left side, such that:
`2sqrt(x^2 + 9x) = -2x - 5`
Squaring again both sides, yields:
`4(x^2 + 9x) = (-(2x + 5))^2`
`4(x^2 + 9x) = 4x^2 + 20x + 25`
`4x^2 + 36x = 4x^2 + 20x + 25`
Reducing duplicate terms both sides yields:
`36x = 20x + 25 => 36x - 20x = 25 => 16x = 25 => x = 25/16`
Testing the value in equation, yields:
`sqrt(25/16) + sqrt(25/16 + 9) = 2`
`5/4 + sqrt(34)/4 = 2 => sqrt(34)/4 = 2 - 5/4`
`sqrt(34)/4 = 3/4`
Since the numerators are not equal `sqrt(34) != 3` , hence, the values `sqrt34/4 != 3/4` , yields that `x =25/16` cannot be considered the solution to the given equation.
Hence, testing if the equation has solution, yields that the given equation has no solution.
Let us examine if this problem has solution or not.
Define a function
Let us restrict our self that f is real function.So domain of f is real number .It means `x >=0`
Let find the zero of f(x) . If zero of f(x) exist then root of the corresponding equation f(x)=0 will also exist.
Graph is moving in an upwrd direction or we can say f(x) is an increasing function in `[0,oo)` . It does not intersect x-axis ,so zero of f(x) does not exist. Thus corresponding equation does not posses root.