Let's solve the equation:

$\sqrt{x} + 2 + 4 = x$

Step 1: Simplify the constants.

$\sqrt{x} + 6 = x$

Step 2: Isolate the square root term.

$\sqrt{x} = x - 6$

Step 3: Square both sides to eliminate the square root.

$(\sqrt{x})^2 = (x - 6)^2$ $x = (x - 6)^2$

Step 4: Expand the square on the right-hand side.

$x = (x - 6)(x - 6) = x^2 - 12x + 36$

Step 5: Rearrange the equation.

$x - x^2 + 12x - 36 = 0$ $-x^2 + 13x - 36 = 0$ Multiply by $-1$ to simplify: $x^2 - 13x + 36 = 0$

Step 6: Solve the quadratic equation using the quadratic formula.

The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $x^2 - 13x + 36 = 0$, the coefficients are:

• $a = 1$
• $b = -13$
• $c = 36$

Substitute into the formula: $x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}$ $x = \frac{13 \pm \sqrt{169 - 144}}{2}$ $x = \frac{13 \pm \sqrt{25}}{2}$ $x = \frac{13 \pm 5}{2}$

Step 7: Find the two possible solutions.

$x_1 = \frac{13 + 5}{2} = \frac{18}{2} = 9$ $x_2 = \frac{13 - 5}{2} = \frac{8}{2} = 4$

Step 8: Check for extraneous solutions.

Substitute $x = 9$ into the original equation: $\sqrt{9} + 6 = 9 \quad \text{which is true since} \quad 3 + 6 = 9$ Substitute $x = 4$ into the original equation: $\sqrt{4} + 6 = 4 \quad \text{which is false since} \quad 2 + 6 \neq 4$

Thus, the only valid solution is: $x = 9$

Do you want any further details or clarifications?

Here are 5 related questions:

1. How do you solve more complex square root equations?
2. What is the quadratic formula, and when is it used?
3. What are extraneous solutions, and why do they occur in radical equations?
4. How can you solve equations with multiple square root terms?
5. What methods are available to check your solution for an equation?

Tip: Always check your solutions when solving radical equations, as squaring both sides can introduce extraneous solutions.