Verified Answer :

  • Determine the domain by ensuring the radicand is non-negative (\(x-3 \geq 0\)) and the denominator is non-zero (\(2x-6 eq 0\)), resulting in the domain \(x > 3\).
  • Find potential roots by setting the factors equal to zero, giving candidates \(x = 3\) and \(x = 7\).
  • Verify the candidates against the domain, finding that \(x = 3\) is excluded due to division by zero, while \(x = 7\) is valid.
  • The valid solution for the equation is \(\boxed{x=7}\).

Explanation

  1. Determining the Domain

We are given the equation \(\frac{x-7}{2x-6} \sqrt{x-3} = 0\). To find the valid solutions for \(x\), we must first determine the domain of this expression, which is the set of all real numbers \(x\) for which the expression is defined.

First, consider the term under the square root, \(\sqrt{x-3}\). For this term to be a real number, the argument must be non-negative:
\[x - 3 \geq 0 \Rightarrow x \geq 3\]

Second, consider the rational expression \(\frac{x-7}{2x-6}\). For this expression to be defined, the denominator must not be zero:
\[2x - 6 eq 0 \Rightarrow 2x eq 6 \Rightarrow x eq 3\]

Combining these two conditions (\(x \geq 3\) and \(x eq 3\)), we determine that the domain of the equation is all real numbers \(x\) strictly greater than 3. Thus, the domain is \((3, \infty)\).

  1. Solving for Potential Roots

Now that we have established the domain, we solve the equation \(\frac{x-7}{2x-6} \sqrt{x-3} = 0\). A product of two terms equals zero if and only if at least one of the factors is equal to zero, provided that the factors are defined within the domain.

We examine each factor:

  1. \(\sqrt{x-3} = 0\): This implies \(x - 3 = 0\), so \(x = 3\).
  2. \(\frac{x-7}{2x-6} = 0\): This implies the numerator \(x - 7 = 0\), so \(x = 7\).
  1. Verification and Final Conclusion

We must verify which of these candidate solutions, \(x=3\) and \(x=7\), lie within our established domain \((3, \infty)\).

  • For \(x = 3\): Our domain restriction from the rational expression requires \(x eq 3\). Substituting \(x=3\) into the original equation leads to a division by zero in the term \(\frac{x-7}{2x-6}\), so \(x=3\) is excluded from the domain.
  • For \(x = 7\): Since \(7 > 3\), this value is in the domain. Substituting \(x=7\) into the equation gives:

\[\frac{7-7}{2(7)-6} \sqrt{7-3} = \frac{0}{8} \sqrt{4} = 0 \cdot 2 = 0\]
This satisfies the equation.

Thus, the only valid solution is \(x = 7\).

Examples

The concept of determining the domain of an equation is crucial in fields like engineering and physics, where variables represent physical quantities. For instance, if you are designing a support structure where the load capacity is described by a function involving a square root of a clearance dimension, you must ensure that your design parameters do not lead to undefined states (like division by zero or imaginary results). Just like in our algebraic problem, identifying the valid range of inputs ensures that your calculations remain grounded in physical reality and that your final results are functional and accurate.

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