Verified Answer :

  • Define the variable \(u = \sqrt{x}\), transforming the radical equation into the quadratic form \(u^2 - u - 6 = 0\).
  • Factor the quadratic equation to obtain \(u = 3\) and \(u = -2\).
  • Apply the constraint \(\sqrt{x} \ge 0\) to reject the negative root, leaving \(u = 3\).
  • Calculate \(x = 3^2 = 9\) and verify it satisfies the original equation \(\boxed{9}\).

Explanation

  1. Analysis and Constraints

We are tasked with solving the radical equation \(\sqrt{x} + 6 = x\) and choosing the correct value from the given options: \(4\), \(9\), \(4\) and \(9\), or no solution. A radical equation involves a variable under a square root sign. For the equation to be defined in real numbers, the term inside the square root must be non-negative, meaning \(x \ge 0\). Additionally, since \(\sqrt{x}\) denotes the principal (non-negative) square root, the equation implies that \(x = \sqrt{x} + 6 \ge 6\).

  1. Substitution and Simplification

To simplify the equation, we perform a substitution. Let \(u = \sqrt{x}\). Given the constraints, we must have \(u \ge 0\). Substituting \(u\) for \(\sqrt{x}\) and noting that \(x = u^2\), the original equation \(\sqrt{x} + 6 = x\) becomes:
\[u + 6 = u^2\]
Rearranging this into a standard quadratic form \(u^2 - u - 6 = 0\), we can solve for \(u\).

  1. Solving the Quadratic Equation

The quadratic equation \(u^2 - u - 6 = 0\) can be factored by finding two numbers that multiply to \(-6\) and add to \(-1\). These numbers are \(-3\) and \(2\), so the equation factors as:
\[(u - 3)(u + 2) = 0\]
This yields two potential solutions for \(u\): \(u = 3\) and \(u = -2\). However, recall our constraint from step 1 that \(u = \sqrt{x} \ge 0\). Therefore, we must discard \(u = -2\) as it is extraneous. We are left with \(u = 3\).

  1. Verification and Final Answer

With the value \(u = 3\), we substitute back to find \(x\):
\[x = u^2 = 3^2 = 9\]
To ensure this is not an extraneous solution, we check it in the original equation \(\sqrt{x} + 6 = x\):
\[\sqrt{9} + 6 = 3 + 6 = 9\]
Since \(9 = 9\), the solution is valid. Testing the value \(x = 4\) (which is one of the options) gives \(\sqrt{4} + 6 = 2 + 6 = 8 eq 4\), confirming that \(4\) is not a solution. Thus, the only correct solution is \(9\).

Examples

Solving radical equations is a foundational skill used in many scientific fields. For example, if an engineer is calculating the distance \(d\) covered by an object under specific acceleration or velocity constraints, they might encounter equations involving square roots of time or displacement. By setting up the equation correctly and checking for potential extraneous results, they can determine the accurate time or distance required for a mechanical component to reach its target state.

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