$$ {\displaylines x-13\sqrt{x}+40=0\\ } $$

We are given the equation:

\[
x - 13\sqrt{x} + 40 = 0
\]

Let's solve it step by step.

### Step 1: Substitution
Let $ y = \sqrt{x} $, which implies $ x = y^2 $. Substituting this into the original equation:

\[
y^2 - 13y + 40 = 0
\]

### Step 2: Solve the quadratic equation
The quadratic equation is now:

\[
y^2 - 13y + 40 = 0
\]

We can solve this using the quadratic formula:

\[
y = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(40)}}{2(1)}
\]

Simplifying:

\[
y = \frac{13 \pm \sqrt{169 - 160}}{2}
\]
\[
y = \frac{13 \pm \sqrt{9}}{2}
\]
\[
y = \frac{13 \pm 3}{2}
\]

So we have two solutions for $ y $:

1. $ y = \frac{13 + 3}{2} = 8 $
2. $ y = \frac{13 - 3}{2} = 5 $

### Step 3: Solve for $ x $
Since $ y = \sqrt{x} $, we can square both solutions for $ y $ to find $ x $:

1. If $ y = 8 $, then $ x = 8^2 = 64 $.
2. If $ y = 5 $, then $ x = 5^2 = 25 $.

### Final solutions:
The solutions to the original equation are:

\[
x = 64 \quad \text{and} \quad x = 25
\]

Would you like more details on any part of the solution?

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Here are 5 related questions to expand on this concept:

1. How do you solve quadratic equations using the quadratic formula?
2. What are the conditions under which substitution works in solving radical equations?
3. How can we check if both solutions are valid in the original equation?
4. What are the differences between solving quadratic equations and cubic equations?
5. How can the discriminant help in determining the nature of solutions?

**Tip:** Always check for extraneous solutions when solving equations involving radicals!

This guide explains how to solve the radical equation x - 13√x + 40 = 0 using substitution and the quadratic formula. Explore step-by-step methods for solving radical and quadratic equations, including using substitution techniques and applying the quadratic formula. Ideal for students in Grades 9-10.

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