Answer :
Sure! To determine which of the given radical equations do not have extraneous solutions, we need to solve each equation step-by-step and check the validity of the solutions. Here are the equations:
1. [tex]\(\sqrt{x+6} = x\)[/tex]
2. [tex]\(x = \sqrt{3x - 2}\)[/tex]
3. [tex]\(\sqrt{-18 - 11x} = x\)[/tex]
4. [tex]\(x = \sqrt{5 - 4x}\)[/tex]
Let's analyze each equation:
### Equation 1: [tex]\(\sqrt{x+6} = x\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x+6})^2 = x^2 \][/tex]
[tex]\[ x + 6 = x^2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ \sqrt{3+6} = 3 \quad \Rightarrow \quad \sqrt{9} = 3 \quad \Rightarrow \quad 3 = 3 \][/tex]
This is valid.
- For [tex]\(x = -2\)[/tex]:
[tex]\[ \sqrt{-2+6} = -2 \quad \Rightarrow \quad \sqrt{4} = -2 \quad \Rightarrow \quad 2 = -2 \][/tex]
This is not valid.
So, the valid solution is [tex]\(x = 3\)[/tex]. The equation [tex]\(\sqrt{x+6} = x\)[/tex] does not have extraneous solutions.
### Equation 2: [tex]\(x = \sqrt{3x - 2}\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ x^2 = 3x - 2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 3x + 2 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x - 2)(x - 1) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x - 2 = 0 \quad \text{or} \quad x - 1 = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 1 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 2 = \sqrt{3(2) - 2} \quad \Rightarrow \quad 2 = \sqrt{6 - 2} \quad \Rightarrow \quad 2 = \sqrt{4} \quad \Rightarrow \quad 2 = 2 \][/tex]
This is valid.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1 = \sqrt{3(1) - 2} \quad \Rightarrow \quad 1 = \sqrt{3 - 2} \quad \Rightarrow \quad 1 = \sqrt{1} \quad \Rightarrow \quad 1 = 1 \][/tex]
This is valid.
So, the valid solutions are [tex]\(x = 2\)[/tex] and [tex]\(x = 1\)[/tex]. The equation [tex]\(x = \sqrt{3x - 2}\)[/tex] does not have extraneous solutions.
### Equation 3: [tex]\(\sqrt{-18 - 11x} = x\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ (\sqrt{-18 - 11x})^2 = x^2 \][/tex]
[tex]\[ -18 - 11x = x^2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 + 11x + 18 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x + 9)(x + 2) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 9 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = -9 \quad \text{or} \quad x = -2 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = -9\)[/tex]:
[tex]\[ \sqrt{-18 - 11(-9)} = -9 \quad \Rightarrow \quad \sqrt{-18 + 99} = -9 \quad \Rightarrow \quad \sqrt{81} = -9 \quad \Rightarrow \quad 9 = -9 \][/tex]
This is not valid.
- For [tex]\(x = -2\)[/tex]:
[tex]\[ \sqrt{-18 - 11(-2)} = -2 \quad \Rightarrow \quad \sqrt{-18 + 22} = -2 \quad \Rightarrow \quad \sqrt{4} = -2 \quad \Rightarrow \quad 2 = -2 \][/tex]
This is not valid.
So, neither solution is valid. The equation [tex]\(\sqrt{-18 - 11x} = x\)[/tex] does have extraneous solutions.
### Equation 4: [tex]\(x = \sqrt{5 - 4x}\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ x^2 = 5 - 4x \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 + 4x - 5 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x + 5)(x - 1) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 5 = 0 \quad \text{or} \quad x - 1 = 0 \][/tex]
[tex]\[ x = -5 \quad \text{or} \quad x = 1 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = -5\)[/tex]:
[tex]\[ -5 = \sqrt{5 - 4(-5)} \quad \Rightarrow \quad -5 = \sqrt{5 + 20} \quad \Rightarrow \quad -5 = \sqrt{25} \quad \Rightarrow \quad -5 = 5 \][/tex]
This is not valid.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1 = \sqrt{5 - 4(1)} \quad \Rightarrow \quad 1 = \sqrt{5 - 4} \quad \Rightarrow \quad 1 = \sqrt{1} \quad \Rightarrow \quad 1 = 1 \][/tex]
This is valid.
So, the valid solution is [tex]\(x = 1\)[/tex]. The equation [tex]\(x = \sqrt{5 - 4x}\)[/tex] does not have extraneous solutions.
### Conclusion:
The equations that do not have extraneous solutions are:
[tex]\[ \sqrt{x + 6} = x \][/tex]
[tex]\[ x = \sqrt{3x - 2} \][/tex]
[tex]\[ x = \sqrt{5 - 4x} \][/tex]
Thus, the correct answer is:
[tex]\[ [1, 2, 4] \][/tex]
This indicates that equations 1, 2, and 4 do not have extraneous solutions.
1. [tex]\(\sqrt{x+6} = x\)[/tex]
2. [tex]\(x = \sqrt{3x - 2}\)[/tex]
3. [tex]\(\sqrt{-18 - 11x} = x\)[/tex]
4. [tex]\(x = \sqrt{5 - 4x}\)[/tex]
Let's analyze each equation:
### Equation 1: [tex]\(\sqrt{x+6} = x\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x+6})^2 = x^2 \][/tex]
[tex]\[ x + 6 = x^2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ \sqrt{3+6} = 3 \quad \Rightarrow \quad \sqrt{9} = 3 \quad \Rightarrow \quad 3 = 3 \][/tex]
This is valid.
- For [tex]\(x = -2\)[/tex]:
[tex]\[ \sqrt{-2+6} = -2 \quad \Rightarrow \quad \sqrt{4} = -2 \quad \Rightarrow \quad 2 = -2 \][/tex]
This is not valid.
So, the valid solution is [tex]\(x = 3\)[/tex]. The equation [tex]\(\sqrt{x+6} = x\)[/tex] does not have extraneous solutions.
### Equation 2: [tex]\(x = \sqrt{3x - 2}\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ x^2 = 3x - 2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 3x + 2 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x - 2)(x - 1) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x - 2 = 0 \quad \text{or} \quad x - 1 = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 1 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 2 = \sqrt{3(2) - 2} \quad \Rightarrow \quad 2 = \sqrt{6 - 2} \quad \Rightarrow \quad 2 = \sqrt{4} \quad \Rightarrow \quad 2 = 2 \][/tex]
This is valid.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1 = \sqrt{3(1) - 2} \quad \Rightarrow \quad 1 = \sqrt{3 - 2} \quad \Rightarrow \quad 1 = \sqrt{1} \quad \Rightarrow \quad 1 = 1 \][/tex]
This is valid.
So, the valid solutions are [tex]\(x = 2\)[/tex] and [tex]\(x = 1\)[/tex]. The equation [tex]\(x = \sqrt{3x - 2}\)[/tex] does not have extraneous solutions.
### Equation 3: [tex]\(\sqrt{-18 - 11x} = x\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ (\sqrt{-18 - 11x})^2 = x^2 \][/tex]
[tex]\[ -18 - 11x = x^2 \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 + 11x + 18 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x + 9)(x + 2) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 9 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
[tex]\[ x = -9 \quad \text{or} \quad x = -2 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = -9\)[/tex]:
[tex]\[ \sqrt{-18 - 11(-9)} = -9 \quad \Rightarrow \quad \sqrt{-18 + 99} = -9 \quad \Rightarrow \quad \sqrt{81} = -9 \quad \Rightarrow \quad 9 = -9 \][/tex]
This is not valid.
- For [tex]\(x = -2\)[/tex]:
[tex]\[ \sqrt{-18 - 11(-2)} = -2 \quad \Rightarrow \quad \sqrt{-18 + 22} = -2 \quad \Rightarrow \quad \sqrt{4} = -2 \quad \Rightarrow \quad 2 = -2 \][/tex]
This is not valid.
So, neither solution is valid. The equation [tex]\(\sqrt{-18 - 11x} = x\)[/tex] does have extraneous solutions.
### Equation 4: [tex]\(x = \sqrt{5 - 4x}\)[/tex]
Step 1: Square both sides to eliminate the square root:
[tex]\[ x^2 = 5 - 4x \][/tex]
Step 2: Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 + 4x - 5 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ (x + 5)(x - 1) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 5 = 0 \quad \text{or} \quad x - 1 = 0 \][/tex]
[tex]\[ x = -5 \quad \text{or} \quad x = 1 \][/tex]
Step 5: Check the solutions in the original equation to ensure there are no extraneous solutions:
- For [tex]\(x = -5\)[/tex]:
[tex]\[ -5 = \sqrt{5 - 4(-5)} \quad \Rightarrow \quad -5 = \sqrt{5 + 20} \quad \Rightarrow \quad -5 = \sqrt{25} \quad \Rightarrow \quad -5 = 5 \][/tex]
This is not valid.
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1 = \sqrt{5 - 4(1)} \quad \Rightarrow \quad 1 = \sqrt{5 - 4} \quad \Rightarrow \quad 1 = \sqrt{1} \quad \Rightarrow \quad 1 = 1 \][/tex]
This is valid.
So, the valid solution is [tex]\(x = 1\)[/tex]. The equation [tex]\(x = \sqrt{5 - 4x}\)[/tex] does not have extraneous solutions.
### Conclusion:
The equations that do not have extraneous solutions are:
[tex]\[ \sqrt{x + 6} = x \][/tex]
[tex]\[ x = \sqrt{3x - 2} \][/tex]
[tex]\[ x = \sqrt{5 - 4x} \][/tex]
Thus, the correct answer is:
[tex]\[ [1, 2, 4] \][/tex]
This indicates that equations 1, 2, and 4 do not have extraneous solutions.
