Verified Answer :

  • Determine the domain constraint \(t \ge 0\) because the square root is non-negative.
  • Square both sides to obtain the quadratic equation \(t^2 - t - 90 = 0\).
  • Factor the quadratic as \((t-10)(t+9) = 0\) to find potential solutions \(t = 10\) and \(t = -9\).
  • Check the solutions and discard the extraneous value \(-9\) to conclude the solution is \(t = 10\).
  • The final solution is \(\boxed{10}\).

Explanation

  1. Analysis of the Radical Equation

We are given the radical equation \(\sqrt{t+90} = t\). To solve for \(t\), we first identify the necessary conditions. Since the square root \(\sqrt{t+90}\) is defined only for \(t+90 \ge 0\), we must have \(t \ge -90\). Furthermore, because the principal square root always outputs a non-negative value, the left side is non-negative, which implies the right side, \(t\), must also be non-negative: \(t \ge 0\). Thus, any valid solution must satisfy \(t \ge 0\).

  1. Transforming into a Quadratic Equation

To eliminate the radical, we square both sides of the equation:
\[(\sqrt{t+90})^2 = t^2\]
\[t+90 = t^2\]
Rearranging the terms to form a standard quadratic equation, we subtract \(t+90\) from both sides:
\[t^2 - t - 90 = 0\]

  1. Solving the Quadratic Equation

We factor the quadratic expression \(t^2 - t - 90 = 0\). We look for two numbers that multiply to \(-90\) and add to \(-1\). These numbers are \(-10\) and \(9\). Thus, the equation can be written as:
\[(t - 10)(t + 9) = 0\]
This gives us two potential solutions:
\[t = 10 \quad \text{or} \quad t = -9\]

  1. Verification and Final Solution

We must verify the solutions against the condition \(t \ge 0\) established in step 1.
For \(t = 10\): \(\sqrt{10+90} = \sqrt{100} = 10\). This satisfies the original equation.
For \(t = -9\): \(\sqrt{-9+90} = \sqrt{81} = 9\). However, the original equation states \(\sqrt{t+90} = t\), and \(9 eq -9\). Therefore, \(t = -9\) is an extraneous solution and is discarded. The only valid solution is \(t = 10\).

Examples

Radical equations appear in many fields, including physics and engineering. For example, if you are calculating the time \(t\) it takes for an object to fall from a certain height, the formula often involves a square root of the distance. Similarly, when determining the speed of an object based on its kinetic energy or velocity-displacement relationships, you frequently encounter radical terms that need to be solved to find the physical values. Solving these correctly is crucial for accurate measurements and predictions in real-world applications.

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