Question
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Are there any extraneous solutions to the following equation? Show how you determine whether there are any extraneous solutions or not.
−3−2x−−−−−−−√=2−x−−−−√
Are there any extraneous solutions to the following equation? Show how you determine whether there are any extraneous solutions or not.
−3−2x−−−−−−−√=2−x−−−−√
Answers
GPT-4o mini
To determine if there are any extraneous solutions for the equation \(-3 - 2\sqrt{-x} = 2 - \sqrt{-x}\), we will follow a systematic approach.
### Step 1: Rearrange the Equation
Start by isolating the radical terms on one side:
\[
-3 - 2\sqrt{-x} = 2 - \sqrt{-x}
\]
Add \(\sqrt{-x}\) to both sides:
\[
-3 - 2\sqrt{-x} + \sqrt{-x} = 2
\]
This simplifies to:
\[
-3 - \sqrt{-x} = 2
\]
Next, add 3 to both sides:
\[
-\sqrt{-x} = 5
\]
### Step 2: Solve for \(\sqrt{-x}\)
Multiply both sides by -1:
\[
\sqrt{-x} = -5
\]
### Step 3: Square Both Sides
We will now square both sides to eliminate the square root:
\[
(-x) = (-5)^2
\]
This leads to:
\[
-x = 25
\]
Next, solve for \(x\):
\[
x = -25
\]
### Step 4: Check for Extraneous Solutions
Now that we have \(x = -25\), we need to check whether this solution satisfies the original equation. Substitute \(x = -25\) back into the original equation:
\[
-3 - 2\sqrt{-(-25)} = 2 - \sqrt{-(-25)}
\]
Calculate \(-(-25)\):
\[
-3 - 2\sqrt{25} = 2 - \sqrt{25}
\]
Since \(\sqrt{25} = 5\), we substitute:
\[
-3 - 2(5) = 2 - 5
\]
This becomes:
\[
-3 - 10 = 2 - 5
\]
Calculating both sides:
Left side:
\[
-3 - 10 = -13
\]
Right side:
\[
2 - 5 = -3
\]
### Step 5: Compare Results
The left side gives \(-13\) and the right side gives \(-3\). Since \(-13 \neq -3\), \(x = -25\) does not satisfy the original equation.
### Conclusion
Thus, the solution \(x = -25\) is extraneous because it does not satisfy the original equation. Therefore, there are extraneous solutions for the equation \( -3 - 2\sqrt{-x} = 2 - \sqrt{-x} \).
### Step 1: Rearrange the Equation
Start by isolating the radical terms on one side:
\[
-3 - 2\sqrt{-x} = 2 - \sqrt{-x}
\]
Add \(\sqrt{-x}\) to both sides:
\[
-3 - 2\sqrt{-x} + \sqrt{-x} = 2
\]
This simplifies to:
\[
-3 - \sqrt{-x} = 2
\]
Next, add 3 to both sides:
\[
-\sqrt{-x} = 5
\]
### Step 2: Solve for \(\sqrt{-x}\)
Multiply both sides by -1:
\[
\sqrt{-x} = -5
\]
### Step 3: Square Both Sides
We will now square both sides to eliminate the square root:
\[
(-x) = (-5)^2
\]
This leads to:
\[
-x = 25
\]
Next, solve for \(x\):
\[
x = -25
\]
### Step 4: Check for Extraneous Solutions
Now that we have \(x = -25\), we need to check whether this solution satisfies the original equation. Substitute \(x = -25\) back into the original equation:
\[
-3 - 2\sqrt{-(-25)} = 2 - \sqrt{-(-25)}
\]
Calculate \(-(-25)\):
\[
-3 - 2\sqrt{25} = 2 - \sqrt{25}
\]
Since \(\sqrt{25} = 5\), we substitute:
\[
-3 - 2(5) = 2 - 5
\]
This becomes:
\[
-3 - 10 = 2 - 5
\]
Calculating both sides:
Left side:
\[
-3 - 10 = -13
\]
Right side:
\[
2 - 5 = -3
\]
### Step 5: Compare Results
The left side gives \(-13\) and the right side gives \(-3\). Since \(-13 \neq -3\), \(x = -25\) does not satisfy the original equation.
### Conclusion
Thus, the solution \(x = -25\) is extraneous because it does not satisfy the original equation. Therefore, there are extraneous solutions for the equation \( -3 - 2\sqrt{-x} = 2 - \sqrt{-x} \).