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Which of the following is an extraneous solution of [tex]\((45 - 3x)^{\frac{1}{2}} = x - 9\)[/tex]?

A. [tex]\(x = -12\)[/tex]
B. [tex]\(x = -3\)[/tex]
C. [tex]\(x = 3\)[/tex]
D. [tex]\(x = 12\)[/tex]


Sagot :

To determine which of the given values [tex]\(-12\)[/tex], [tex]\(-3\)[/tex], [tex]\(3\)[/tex], and [tex]\(12\)[/tex] are extraneous solutions to the equation [tex]\((45 - 3x)^{\frac{1}{2}} = x - 9\)[/tex], we will evaluate each one in the context of the original equation. An extraneous solution is one that does not satisfy the original equation when substituted back in.

### Step-by-Step Solution:

1. Original equation:
[tex]\[ (45 - 3x)^{\frac{1}{2}} = x - 9 \][/tex]

2. Check [tex]\(x = -12\)[/tex]:
[tex]\[ \text{Left-hand side (LHS)}: (45 - 3(-12))^{\frac{1}{2}} = (45 + 36)^{\frac{1}{2}} = 81^{\frac{1}{2}} = 9 \][/tex]
[tex]\[ \text{Right-hand side (RHS)}: -12 - 9 = -21 \][/tex]
[tex]\[ \text{Since \(9 \neq -21\), \(x = -12\) is an extraneous solution.} \][/tex]

3. Check [tex]\(x = -3\)[/tex]:
[tex]\[ \text{Left-hand side (LHS)}: (45 - 3(-3))^{\frac{1}{2}} = (45 + 9)^{\frac{1}{2}} = 54^{\frac{1}{2}} = \sqrt{54} \][/tex]
[tex]\[ \text{Right-hand side (RHS)}: -3 - 9 = -12 \][/tex]
[tex]\[ \text{Since \(\sqrt{54} \neq -12\), \(x = -3\) is an extraneous solution.} \][/tex]

4. Check [tex]\(x = 3\)[/tex]:
[tex]\[ \text{Left-hand side (LHS)}: (45 - 3(3))^{\frac{1}{2}} = (45 - 9)^{\frac{1}{2}} = 36^{\frac{1}{2}} = 6 \][/tex]
[tex]\[ \text{Right-hand side (RHS)}: 3 - 9 = -6 \][/tex]
[tex]\[ \text{Since \(6 \neq -6\), \(x = 3\) is an extraneous solution.} \][/tex]

5. Check [tex]\(x = 12\)[/tex]:
[tex]\[ \text{Left-hand side (LHS)}: (45 - 3(12))^{\frac{1}{2}} = (45 - 36)^{\frac{1}{2}} = 9^{\frac{1}{2}} = 3 \][/tex]
[tex]\[ \text{Right-hand side (RHS)}: 12 - 9 = 3 \][/tex]
[tex]\[ \text{Since \(3 = 3\), \(x = 12\) is \textbf{not} an extraneous solution.} \][/tex]

### Conclusion:
Based on the evaluations above, the extraneous solutions to the equation [tex]\((45 - 3x)^{\frac{1}{2}} = x - 9\)[/tex] are:
- [tex]\(x = -12\)[/tex]
- [tex]\(x = -3\)[/tex]
- [tex]\(x = 3\)[/tex]

Therefore, the extraneous solutions are [tex]\(-12\)[/tex], [tex]\(-3\)[/tex], and [tex]\(3\)[/tex].