Solving an irrational equation

[Mr Davis 97]

I have the irrational equation $\sqrt{x - 1} + \sqrt{2 - x} = 0$, which has no real solutions. However, when I try to solve the equation, I get a real solution, that is:

$\sqrt{x - 1} + \sqrt{2 - x} = 0$
$\sqrt{x - 1} = -\sqrt{2 - x}$
$(\sqrt{x - 1})^{2} = (-\sqrt{2 - x})^{2}$
$(\sqrt{x - 1})(\sqrt{x - 1}) = (-\sqrt{2 - x})(-\sqrt{2 - x})$
$x - 1 = 2 - x$
$x = \frac{3}{2}$

What am I doing wrong here? In which step am I making a mistake?

 

[Mark44]

I have the irrational equation $\sqrt{x - 1} + \sqrt{2 - x} = 0$, which has no real solutions. However, when I try to solve the equation, I get a real solution, that is:

$\sqrt{x - 1} + \sqrt{2 - x} = 0$
$\sqrt{x - 1} = -\sqrt{2 - x}$
$(\sqrt{x - 1})^{2} = (-\sqrt{2 - x})^{2}$
$(\sqrt{x - 1})(\sqrt{x - 1}) = (-\sqrt{2 - x})(-\sqrt{2 - x})$
$x - 1 = 2 - x$
$x = \frac{3}{2}$

What am I doing wrong here? In which step am I making a mistake?

When you square both sides of an equation, there's the possibility of getting an extraneous solution, one that isn't a solution of the original equation. Here's a simple example.

x = -2 -- solution set is {-2}
Square both sides:
x² = (-2)² = 4 -- solution set is now {-2, 2}

By squaring both sides, the solution set went from {-2} to {-2, 2}. As far as the original equation is concerned, the solution x = 2 is extraneous.

The bottom line is: Whenever you square both sides of an equation, you need to check that any solution you get satisifies the original equation.

 

[symbolipoint]

I have the irrational equation $\sqrt{x - 1} + \sqrt{2 - x} = 0$, which has no real solutions. However, when I try to solve the equation, I get a real solution, that is:

$\sqrt{x - 1} + \sqrt{2 - x} = 0$
$\sqrt{x - 1} = -\sqrt{2 - x}$
$(\sqrt{x - 1})^{2} = (-\sqrt{2 - x})^{2}$
$(\sqrt{x - 1})(\sqrt{x - 1}) = (-\sqrt{2 - x})(-\sqrt{2 - x})$
$x - 1 = 2 - x$
$x = \frac{3}{2}$

What am I doing wrong here? In which step am I making a mistake?

Checking your found value in the original equation, decide if that value works or not.