To solve the equation $\sqrt{12 - x} = x$, let's proceed step by step.

1. Square both sides to eliminate the square root: $(\sqrt{12 - x})^2 = x^2$ This simplifies to: $12 - x = x^2$

2. Rearrange the equation to form a quadratic equation: $x^2 + x - 12 = 0$

3. 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 $x^2 + x - 12 = 0$, the coefficients are $a = 1$, $b = 1$, and $c = -12$.

   Substituting these into the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 48}}{2}$ $x = \frac{-1 \pm \sqrt{49}}{2}$ $x = \frac{-1 \pm 7}{2}$

4. Calculate the two possible solutions:

   • For the positive root: $x = \frac{-1 + 7}{2} = \frac{6}{2} = 3$
   • For the negative root: $x = \frac{-1 - 7}{2} = \frac{-8}{2} = -4$

5. Check for extraneous solutions: Since we squared both sides of the equation earlier, we need to check both solutions to make sure they satisfy the original equation.

   • For $x = 3$: $\sqrt{12 - 3} = \sqrt{9} = 3$ This is true, so $x = 3$ is a valid solution.

   • For $x = -4$: $\sqrt{12 - (-4)} = \sqrt{12 + 4} = \sqrt{16} = 4$ But the original equation was $\sqrt{12 - x} = x$, and this does not hold true since $4 \neq -4$. So, $x = -4$ is an extraneous solution.

Final Answer:

The only valid solution is: $x = 3$

Would you like any further clarification or additional steps? Here are some related questions for you:

1. How do you check for extraneous solutions in square root equations?
2. What other methods can be used to solve quadratic equations?
3. What happens if you don't check for extraneous solutions after squaring both sides?
4. How can you solve square root equations that have multiple terms under the radical?
5. Can you solve a quadratic equation by factoring instead of using the quadratic formula?

Tip: When solving square root equations, always remember to check your solutions in the original equation, especially after squaring both sides, as this may introduce extraneous solutions.