Solving equation involing radicals

[issacnewton]

Homework Statement

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

where [itex]b>a>0[/itex]

Homework Equations

equation is given above

The Attempt at a Solution

I squared both sides and and solved this. I got two solutions [itex]x=a[/itex] and [itex]x=b[/itex]. 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 [itex]x=a[/itex] 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 [itex]x=b[/itex] in the original equation, I get the following [itex]\sqrt{a-b} = \sqrt{a-b}[/itex]. Now here left side matches with the right side. But since [itex]b>a>0[/itex], both sides are not real number anymore, So is [itex]x=b[/itex] extraneous solution or is it the second possible solution ?

thanks

 

[Mark44]

Homework Statement

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

where [itex]b>a>0[/itex]

Homework Equations

equation is given above

The Attempt at a Solution

I squared both sides and and solved this. I got two solutions [itex]x=a[/itex] and [itex]x=b[/itex]. 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 [itex]x=a[/itex] 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 [itex]x=b[/itex] in the original equation, I get the following [itex]\sqrt{a-b} = \sqrt{a-b}[/itex]. Now here left side matches with the right side. But since [itex]b>a>0[/itex], both sides are not real number anymore, So is [itex]x=b[/itex] extraneous solution or is it the second possible solution ?

thanks

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

 

[issacnewton]

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

 

[Ray Vickson]

But Mark, left side matches with the right side in the case of [itex]x=b[/itex]. 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:
[tex] \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
[/tex]

 

[issacnewton]

Ray, so is [itex]x=b[/itex] an extraneous solution or not ?

 

[Mark44]

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.

 

[issacnewton]

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

 

[BruceW]

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.

 

[issacnewton]

thanks Bruce