Let's solve the problems one by one.

Problem 13

Given: $x - \sqrt{\frac{20}{x}} = 7$ You need to find the value of $\sqrt{5x} - x$.

Problem 14

Given: $x - 2 = \sqrt{x}$ You need to find the value of $x + \frac{4}{x}$.

Let's start solving these.

Problem 13 Solution:

1. Given equation: $x - \sqrt{\frac{20}{x}} = 7$ Rearrange the equation to isolate the square root term: $\sqrt{\frac{20}{x}} = x - 7$

2. Square both sides to eliminate the square root: $\frac{20}{x} = (x - 7)^2$ Expand the right side: $\frac{20}{x} = x^2 - 14x + 49$

3. Multiply both sides by $x$ to get rid of the fraction: $20 = x^3 - 14x^2 + 49x$ This is a cubic equation in $x$, which can be solved using various methods, such as factoring or using a numerical solver.

4. Assume a solution, and once you have $x$, substitute it back to find the value of $\sqrt{5x} - x$.

Problem 14 Solution:

1. Given equation: $x - 2 = \sqrt{x}$ Rearrange the equation: $x = \sqrt{x} + 2$

2. Square both sides to eliminate the square root: $x^2 = x + 4$ Rearrange to form a quadratic equation: $x^2 - x - 4 = 0$

3. Solve the quadratic equation using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)}$ Simplify: $x = \frac{1 \pm \sqrt{17}}{2}$ Select the appropriate solution for $x$ (usually the positive one).

4. Once you have $x$, substitute it into $x + \frac{4}{x}$ to find the desired value.

Would you like more detailed steps on how to solve the cubic or quadratic equation, or how to proceed with the rest of the problem?

Related Questions:

1. What is the general method to solve a cubic equation?
2. How can you factor a quadratic equation?
3. What is the significance of squaring both sides of an equation when solving it?
4. How would you approach solving an equation with a square root in a more complex scenario?
5. Can you solve systems of equations that involve both quadratic and cubic terms?

Tip:

When dealing with equations that have square roots, always remember to check for extraneous solutions after squaring both sides, as this process can introduce solutions that don't satisfy the original equation.