Why doesn't this radical equation have a solution?

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SUMMARY

The equation √(x-5) - √(x-8) = 3 has no solution due to the properties of square roots. The attempt to solve the equation led to the conclusion that -1 = √(x-8), which is impossible since the square root function only yields non-negative results. Although squaring both sides produced x = 9, substituting this value back into the original equation results in a false statement, confirming that x = 9 is an extraneous solution.

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ryn droma
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I don't understand why this is a no solution



original problem
√(x-5) - √(x-8) = 3




My solution attempt

add +√(x-8) to both sides:
√(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
 
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ryn droma said:
I don't understand why this is a no solution



original problem
√(x-5) - √(x-8) = 3




My solution attempt

add +√(x-8) to both sides:
√(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?
Correct.
ryn droma said:
I continued with squaring both sides again:
(-1)^2 = √(x-8)^2

1 = x - 8
+8

x = 9

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.
 
welcome to pf!

hi ryn droma! welcome to pf! :smile:
ryn droma said:
√(x-5) - √(x-8) = 3

-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

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 :wink:)
 

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