# Verify if equation sq root x+ sqroot(x+9)=2 have roots?

embizze | High School Teacher | (Level 2) Educator Emeritus

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Determine if `sqrt(x)+sqrt(x+9)=2` has any solutions:

`x>=0` because the domain of `sqrt(x)` is nonnegative x.

`sqrt(x)>=0`

`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.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.

aruv | High School Teacher | (Level 2) Valedictorian

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Let us examine if this problem has solution or not.

Define a function

`f(x)=sqrt(x)+sqrt(x+9)-2`

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) isĀ 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.