Verified Answer :

  • Establish domain constraints \(4 \leq z \leq 8.8\) for the square root to yield a real, valid result.
  • Square both sides to eliminate the radical and rearrange into the quadratic form \(z^2 - 3z - 28 = 0\).
  • Apply the quadratic formula to find the potential solutions \(z = 7\) and \(z = -4\).
  • Validate the solutions against the domain constraints, concluding that the only real solution is \(\boxed{7}\).

Explanation

  1. Analysis and Domain Conditions

To solve the radical equation \(\sqrt{44-5z} = z-4\), we must first identify the domain of validity for the equation.

  1. The expression inside the square root must be non-negative:

\[44-5z \geq 0 \implies 5z \leq 44 \implies z \leq 8.8\]

  1. The square root of a real number is always non-negative, so the right side of the equation must also be non-negative:

\[z-4 \geq 0 \implies z \geq 4\]
Thus, any valid solution for \(z\) must satisfy the condition \(4 \leq z \leq 8.8\).

  1. Squaring and Simplifying the Equation

To solve for \(z\), we square both sides of the original equation:
\[(\sqrt{44-5z})^2 = (z-4)^2\]
\[44-5z = z^2 - 8z + 16\]
Rearranging the terms to form a standard quadratic equation:
\[z^2 - 8z + 5z + 16 - 44 = 0\]
\[z^2 - 3z - 28 = 0\]

  1. Solving the Quadratic Equation

We can solve the quadratic equation \(z^2 - 3z - 28 = 0\) using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[z = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-28)}}{2(1)}\]
\[z = \frac{3 \pm \sqrt{9 + 112}}{2}\]
\[z = \frac{3 \pm \sqrt{121}}{2}\]
\[z = \frac{3 \pm 11}{2}\]
The two potential solutions are:

  1. \(z_1 = \frac{3+11}{2} = \frac{14}{2} = 7\)
  2. \(z_2 = \frac{3-11}{2} = \frac{-8}{2} = -4\)
  1. Verification and Final Result

Finally, we must verify which of these candidates satisfy the domain condition \(4 \leq z \leq 8.8\):

  • For \(z = 7\): \(4 \leq 7 \leq 8.8\). This condition is satisfied. Plugging it back into the original equation, we get \(\sqrt{44-5(7)} = \sqrt{9} = 3\), and \(7-4 = 3\). Since \(3=3\), \(z=7\) is a valid solution.
  • For \(z = -4\): \(-4\) is not greater than or equal to 4. Therefore, it is an extraneous solution. Plugging it into the original equation, we get \(\sqrt{44-5(-4)} = \sqrt{64} = 8\), but \(-4-4 = -8\). Since \(8 eq -8\), \(z=-4\) is not a solution.

The only valid real solution is \(z = 7\).

Examples

This radical equation serves as a simplified model for physical problems where quantities must be positive to exist, such as calculating the speed of an object or the distance covered under specific constraints. For instance, if a rocket's altitude is governed by a square root function based on fuel consumption, engineers must ensure the resulting altitude is a physically meaningful, positive value. Knowing how to solve these equations and verify against constraints ensures that calculations yield accurate and applicable results in engineering and science.

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