squareroot27-x =x+3 solve for extraneous solution

We are given the equation:

\[
\sqrt{27 - x} = x + 3
\]

### Steps to Solve:

1. **Square both sides** to eliminate the square root:
   \[
   (\sqrt{27 - x})^2 = (x + 3)^2
   \]
   \[
   27 - x = (x + 3)^2
   \]
   Expand the right-hand side:
   \[
   27 - x = x^2 + 6x + 9
   \]

2. **Move all terms to one side** to form a quadratic equation:
   \[
   27 - x - (x^2 + 6x + 9) = 0
   \]
   Simplify:
   \[
   27 - x - x^2 - 6x - 9 = 0
   \]
   Combine like terms:
   \[
   -x^2 - 7x + 18 = 0
   \]
   Multiply through by \(-1\) to make the equation easier to solve:
   \[
   x^2 + 7x - 18 = 0
   \]

3. **Solve the quadratic equation** using the quadratic formula:
   The quadratic formula is:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For the equation \(x^2 + 7x - 18 = 0\), \(a = 1\), \(b = 7\), and \(c = -18\).
   Substituting these values into the quadratic formula:
   \[
   x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-18)}}{2(1)}
   \]
   \[
   x = \frac{-7 \pm \sqrt{49 + 72}}{2}
   \]
   \[
   x = \frac{-7 \pm \sqrt{121}}{2}
   \]
   \[
   x = \frac{-7 \pm 11}{2}
   \]

   So, the two possible solutions are:
   \[
   x = \frac{-7 + 11}{2} = \frac{4}{2} = 2
   \]
   \[
   x = \frac{-7 - 11}{2} = \frac{-18}{2} = -9
   \]

4. **Check for extraneous solutions** by substituting the solutions back into the original equation.

   - For \(x = 2\):
     \[
     \sqrt{27 - 2} = 2 + 3
     \]
     \[
     \sqrt{25} = 5
     \]
     This is true.

   - For \(x = -9\):
     \[
     \sqrt{27 - (-9)} = -9 + 3
     \]
     \[
     \sqrt{36} = -6
     \]
     But \(\sqrt{36} = 6\), not \(-6\), so this is **false**.

### Conclusion:
The solution \(x = 2\) is valid, while \(x = -9\) is an **extraneous solution**.

Do you want more details or have any questions?

Here are 5 related questions:
1. How can you identify extraneous solutions when solving radical equations?
2. What is the significance of squaring both sides in an equation?
3. Can all quadratic equations be solved using the quadratic formula?
4. Why does squaring sometimes introduce extraneous solutions?
5. How do you solve equations involving cube roots?

**Tip:** Always check your solutions by substituting them back into the original equation when dealing with square roots to avoid extraneous solutions.

Learn how to solve the radical equation √(27 - x) = x + 3 and identify extraneous solutions with a detailed step-by-step explanation. This solution involves squaring both sides, forming a quadratic equation, and checking for extraneous solutions. Suitable for students in Grades 9-11.

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