# Homework Help: Why doesn't this radical equation have a solution?

1. Jun 3, 2013

### ryn droma

I dont understand why this is a no solution

original problem
√(x-5) - √(x-8) = 3

My solution attempt

√(x-5) - √(x-8) = 3

√(x-5) = 3 + √(x-8)

squaring both sides:
(√(x-5))^2 = (3 + √(x-8))^2

FOIL out the right side:
x - 5 = 3^2 + 2(3√(x-8)) + √(x-8)^2

simplify:
x - 5 = 9 + 6√(x-8) + x - 8

combine like terms:
x - 5 = x + 1 + 6√(x-8)

Subtract x and 1 from both sides:
x - 5 = x + 1 + 6√(x-8)

-6 = 6√(x-8)

divide both sides by 6:
-1 = √(x-8)

At this point, is it the -1 that means no solution?

I continued with squaring both sides again:
(-1)^2 = √(x-8)^2

1 = x - 8
+8

x = 9

2. Jun 3, 2013

### Staff: Mentor

Correct.
A simpler problem is this:
√w - √(w - 3) = 3
If you sketch the graphs of y = √w and y = √(w - 3), you will see that there are no values of w for which the first graph is 3 units above the second graph. That's why there is no solution to this equation or to the one you posted.

3. Jun 3, 2013

### tiny-tim

welcome to pf!

hi ryn droma! welcome to pf!
it's because of the definition of √

√ always means the positive square root

if you put x = 9 into the original equation, you get

√(9 - 5) - √(9 - 8) = 3, which is false

if you allow negative square roots, then

√(9 - 5) - √(9 - 8) = 2 - (-1) = 3, which is true​

(every time you square an equation, you get extra solutions …

x = 9 is your extra solution )