Why doesn't this radical equation have a solution?

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In summary, the original problem is √(x-5) - √(x-8) = 3 and the solution attempt involved adding √(x-8) to both sides, squaring both sides, FOILing, simplifying, and combining like terms to get -1 = √(x-8), indicating that there is no solution to the problem. This is because the definition of √ always means the positive square root, so when considering the possible solutions, x = 9 is the extra solution obtained when squaring both sides.
  • #1
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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  • #2
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.
 
  • #3
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:)
 

Related to Why doesn't this radical equation have a solution?

1. Why can't I solve this radical equation?

There are a few reasons why a radical equation may not have a solution. One possibility is that the equation is not solvable using traditional algebraic methods. Another possibility is that the equation has no real solutions, meaning the solutions involve imaginary numbers.

2. How do I know if a radical equation has no solution?

If you are unable to isolate the variable and solve for it, or if you end up with a negative number under the radical, then the equation has no solution. You can also graph the equation to see if it intersects with the x-axis at any point. If it does not, then there are no real solutions.

3. Can I use the quadratic formula to solve a radical equation?

The quadratic formula can only be used to solve equations in the form of ax^2 + bx + c = 0. If your radical equation does not fit this form, then the quadratic formula cannot be used.

4. Is there a way to simplify a radical equation before solving it?

Yes, you can simplify a radical equation by factoring out perfect square factors from the radicand. This will make it easier to isolate the variable and solve for it. However, it is not always possible to simplify a radical equation before solving it.

5. Can a radical equation have more than one solution?

Yes, it is possible for a radical equation to have more than one solution. This can happen when the equation has multiple perfect square factors that can be simplified. However, it is also possible for a radical equation to have no solution or only one solution.

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