Answered

Find the real solution of the following radical equation.



$\begin{array}{c}

\sqrt{4 n+12}=n \\

n=[?]

\end{array}$

Verified Answer :

  • Square both sides to eliminate the radical: \(4n + 12 = n^2\).
  • Rearrange into a quadratic equation: \(n^2 - 4n - 12 = 0\).
  • Factor the equation to find potential solutions: \((n - 6)(n + 2) = 0\), giving \(n=6\) and \(n=-2\).
  • Verify the roots: \(n=6\) is valid, while \(n=-2\) is extraneous because it results in a negative value for the square root output.
  • Final answer: \(n = \boxed{6}\).

Explanation

  1. Problem Analysis and Constraints

The goal is to solve the radical equation \(\sqrt{4n + 12} = n\). First, we identify the domain of the equation. For the expression under the square root to be a real number, we must have \(4n + 12 \ge 0\), which simplifies to \(n \ge -3\). Furthermore, since the principal square root \(\sqrt{x}\) always produces a non-negative value, the equation \(\sqrt{4n + 12} = n\) implies that \(n \ge 0\). Consequently, any valid solution for \(n\) must satisfy \(n \ge 0\).

  1. Simplifying the Radical Equation

To eliminate the square root, we square both sides of the equation:
\[(\sqrt{4n + 12})^2 = n^2\]
This simplifies to:
\[4n + 12 = n^2\]
Now, we rearrange the equation to form a standard quadratic equation:
\[n^2 - 4n - 12 = 0\]

  1. Solving the Quadratic Equation

We solve the quadratic equation \(n^2 - 4n - 12 = 0\) by factoring. We look for two numbers that multiply to \(-12\) and add up to \(-4\). These numbers are \(-6\) and \(2\). Thus, the equation can be factored as:
\[(n - 6)(n + 2) = 0\]
Setting each factor to zero, we obtain two potential solutions:
\[n - 6 = 0 \implies n = 6\]
\[n + 2 = 0 \implies n = -2\]

  1. Verification of Solutions

We must verify our potential solutions against the original equation and the constraint \(n \ge 0\).
For \(n = 6\): \(\sqrt{4(6) + 12} = \sqrt{24 + 12} = \sqrt{36} = 6\). This is equal to \(n\), so \(n = 6\) is a valid solution.
For \(n = -2\): \(\sqrt{4(-2) + 12} = \sqrt{-8 + 12} = \sqrt{4} = 2\). However, \(2 eq -2\), and \(n = -2\) violates our constraint \(n \ge 0\). Therefore, \(n = -2\) is an extraneous solution and must be discarded.
Thus, the only real solution is \(n = 6\).

Examples

Radical equations often appear when calculating distances, such as in the Pythagorean theorem, or in physics problems involving velocities. For example, if you know that the time it takes for an object to fall a certain distance is related to the square root of that distance, you might encounter an equation like \(t = \sqrt{h/4.9}\). Solving these equations helps determine unknown variables like initial height or speed in various real-world scenarios.

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