## Homework Statement

Hello , I need to find the real number solutions for the following equation.
$$\sqrt{a-x} + \sqrt{b-x} = \sqrt{a+b-2x}$$

where $b>a>0$

## Homework Equations

equation is given above

## The Attempt at a Solution

I squared both sides and and solved this. I got two solutions $x=a$ and $x=b$. Now when we square both sides of the equations, there is possibility of getting some solutions which may not satisfy the original equation. Such solutions are called extraneous solutions. When I plug in $x=a$ in the original equations, LHS matches with the RHS. So its one of the solution which is a real number. But when I plug in the other possible solution $x=b$ in the original equation, I get the following $\sqrt{a-b} = \sqrt{a-b}$. Now here left side matches with the right side. But since $b>a>0$, both sides are not real number anymore, So is $x=b$ extraneous solution or is it the second possible solution ?

thanks

Mark44
Mentor

## Homework Statement

Hello , I need to find the real number solutions for the following equation.
$$\sqrt{a-x} + \sqrt{b-x} = \sqrt{a+b-2x}$$

where $b>a>0$

## Homework Equations

equation is given above

## The Attempt at a Solution

I squared both sides and and solved this. I got two solutions $x=a$ and $x=b$. Now when we square both sides of the equations, there is possibility of getting some solutions which may not satisfy the original equation. Such solutions are called extraneous solutions. When I plug in $x=a$ in the original equations, LHS matches with the RHS. So its one of the solution which is a real number. But when I plug in the other possible solution $x=b$ in the original equation, I get the following $\sqrt{a-b} = \sqrt{a-b}$. Now here left side matches with the right side. But since $b>a>0$, both sides are not real number anymore, So is $x=b$ extraneous solution or is it the second possible solution ?

thanks
x = b is the extraneous solution for the reason you give.

But Mark, left side matches with the right side in the case of $x=b$. Extraneous solution means that, left side doesn't match with the right side. So I am little confused about my reasoning here.

Ray Vickson
Homework Helper
Dearly Missed
But Mark, left side matches with the right side in the case of $x=b$. Extraneous solution means that, left side doesn't match with the right side. So I am little confused about my reasoning here.

The left and right sides DO match:
$$\sqrt{a-x} + \sqrt{b-x} = \sqrt{a-b} + \sqrt{0} \; \text{when } \; x = b\\ \sqrt{a+b-2x} = \sqrt{a-b} \; \text{when } \; x = b$$

Ray, so is $x=b$ an extraneous solution or not ?

Mark44
Mentor
When x = b, a - b < 0, so ##\sqrt{a - b}## is imaginary. However, you do get a true statement when x = b, and b is a real number, so I guess I'll revise my earlier statement, and say that both a and b are solutions.

Ok thanks Mark. But it seems weird that to get the real solution, the equality ends up involving imaginary numbers....

BruceW
Homework Helper
If the solutions to both expressions (i.e. both sides of the equation) are meant to be real, then x=b would not be allowed. In other words, if we are restricting the domain of possible 'x' values to numbers for which the square root function (in this equation) gives a real number output, then 'b' would not be in that domain. I think it's not totally clear if you are meant to assume this or not.

edit: but most likely, x=b should be a solution too, since the problem statement is a bit vague on this.

thanks Bruce