The problem asks to find the value of the constant $k$ given the equation $\sqrt{3k} + \sqrt{18} - \sqrt{98} = \sqrt{50}$.

1. Problem Description

The problem asks to find the value of the constant $k$ given the equation $\sqrt{3k} + \sqrt{18} - \sqrt{98} = \sqrt{50}$.

2. Solution Steps

First, simplify the square roots:

$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$

$\sqrt{98} = \sqrt{49 \times 2} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}$

$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$

Substitute these simplified values into the equation:

$\sqrt{3k} + 3\sqrt{2} - 7\sqrt{2} = 5\sqrt{2}$

Combine like terms:

$\sqrt{3k} - 4\sqrt{2} = 5\sqrt{2}$

Add $4\sqrt{2}$ to both sides of the equation:

$\sqrt{3k} = 5\sqrt{2} + 4\sqrt{2}$

$\sqrt{3k} = 9\sqrt{2}$

Square both sides of the equation:

$(\sqrt{3k})^2 = (9\sqrt{2})^2$

$3k = 81 \times 2$

$3k = 162$

Divide both sides by 3:

$k = \frac{162}{3}$

$k = 54$

3. Final Answer

The value of the constant $k$ is

5

4.