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The given equation is:

\[
\sqrt{p + 30} = p
\]

To solve for \( p \), let's follow these steps:

### Step 1: Square both sides of the equation
By squaring both sides, we eliminate the square root:

\[
(\sqrt{p + 30})^2 = p^2
\]

This simplifies to:

\[
p + 30 = p^2
\]

### Step 2: Rearrange the equation
Move all terms to one side of the equation to set it equal to zero:

\[
p^2 - p - 30 = 0
\]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \( p^2 - p - 30 = 0 \) using the quadratic formula:

\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( p^2 - p - 30 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -30 \)

Substitute these values into the quadratic formula:

\[
p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-30)}}{2(1)}
\]

\[
p = \frac{1 \pm \sqrt{1 + 120}}{2}
\]

\[
p = \frac{1 \pm \sqrt{121}}{2}
\]

\[
p = \frac{1 \pm 11}{2}
\]

### Step 4: Find the possible values for \( p \)
Now calculate the two possible solutions:

\[
p = \frac{1 + 11}{2} = \frac{12}{2} = 6
\]

\[
p = \frac{1 - 11}{2} = \frac{-10}{2} = -5
\]

### Step 5: Check the solutions
We need to check which of these solutions satisfies the original equation \( \sqrt{p + 30} = p \).

- For \( p = 6 \):

\[
\sqrt{6 + 30} = \sqrt{36} = 6
\]

This is true, so \( p = 6 \) is a valid solution.

- For \( p = -5 \):

\[
\sqrt{-5 + 30} = \sqrt{25} = 5
\]

But \( -5 \neq 5 \), so \( p = -5 \) is not a valid solution.

### Final Answer:
The real solution is:

\[
p = 6
\]

Would you like more details or explanations on any steps? Here are 5 related questions to deepen your understanding:

1. How do you know when to square both sides of an equation?
2. Why is it important to check for extraneous solutions when solving radical equations?
3. How would you solve a more complicated radical equation, like \( \sqrt{2p + 10} = p - 3 \)?
4. Can a quadratic equation always be solved by factoring, or are other methods sometimes necessary?
5. What is the discriminant, and how does it help in solving quadratic equations?

**Tip:** Always check for extraneous solutions when solving radical equations. Squaring both sides can sometimes introduce solutions that don't satisfy the original equation.

Find the real solution of the following radical equation: √p + 30 = p

Learn how to solve the radical equation √p + 30 = p using the quadratic formula. This step-by-step solution involves squaring both sides, rearranging terms, and checking for extraneous solutions. Suitable for students in Grades 9-11.

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