Question

The number of real Solution of the equation \[ \sqrt{x + 4} + \sqrt{5 - x} = 5 \] \text{is:}

• A. 5
• B. 1
• C. 2
• D. 3

Hint:

For radical equations, always check the domain and maximum/minimum values to determine the number of real Solution.

Answer 1

The Correct Option is A

Domain constraints: \[ x + 4 \geq 0 \implies x \geq -4, \quad 5 - x \geq 0 \implies x \leq 5 \] Domain: \([-4, 5]\). Let \[ f(x) = \sqrt{x + 4} + \sqrt{5 - x} \] Check max value of \(f(x)\): At \(x = -4\): \[ f(-4) = 0 + \sqrt{9} = 3 \] At \(x = 5\): \[ f(5) = \sqrt{9} + 0 = 3 \] Find critical point by differentiating \(f(x)\): \[ f'(x) = \frac{1}{2\sqrt{x + 4}} - \frac{1}{2\sqrt{5 - x}} = 0 \implies \sqrt{5 - x} = \sqrt{x + 4} \] Square both sides: \[ 5 - x = x + 4 \implies 2x = 1 \implies x = \frac{1}{2} \] Evaluate \(f\) at \(x = \frac{1}{2}\): \[ f\left(\frac{1}{2}\right) = \sqrt{\frac{1}{2} + 4} + \sqrt{5 - \frac{1}{2}} = \sqrt{4.5} + \sqrt{4.5} = 2 \times \sqrt{4.5} = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2} \approx 4.2426 \] Since max \(f(x) \approx 4.2426<5\), equation \[ \sqrt{x + 4} + \sqrt{5 - x} = 5 \]