Solve the given radical equation. Check all proposed solutions.
StartRoot 2 x plus 20 EndRoot equals x plus 62x+20=x+6

Let's solve the given radical equation step by step:

The given equation is:

\[
\sqrt{2x + 20} = x + 6
\]

### Step 1: Square both sides to eliminate the square root

Squaring both sides of the equation:

\[
\left(\sqrt{2x + 20}\right)^2 = (x + 6)^2
\]

This simplifies to:

\[
2x + 20 = (x + 6)^2
\]

### Step 2: Expand the right-hand side

Now, expand \( (x + 6)^2 \):

\[
2x + 20 = x^2 + 12x + 36
\]

### Step 3: Rearrange the equation

Move all terms to one side to form a quadratic equation:

\[
2x + 20 - 2x - 20 = x^2 + 12x + 36 - 2x - 20
\]

Simplifying:

\[
0 = x^2 + 10x + 16
\]

Or:

\[
x^2 + 10x + 16 = 0
\]

### Step 4: Solve the quadratic equation

We can solve the quadratic equation using the quadratic formula:

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

For the equation \(x^2 + 10x + 16 = 0\), the coefficients are:
- \(a = 1\)
- \(b = 10\)
- \(c = 16\)

Substitute these values into the quadratic formula:

\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(16)}}{2(1)}
\]

\[
x = \frac{-10 \pm \sqrt{100 - 64}}{2}
\]

\[
x = \frac{-10 \pm \sqrt{36}}{2}
\]

\[
x = \frac{-10 \pm 6}{2}
\]

This gives two possible solutions:

\[
x_1 = \frac{-10 + 6}{2} = \frac{-4}{2} = -2
\]
\[
x_2 = \frac{-10 - 6}{2} = \frac{-16}{2} = -8
\]

### Step 5: Check the proposed solutions

We now need to check both solutions in the original equation to ensure they are valid.

#### Checking \(x = -2\):

Substitute \(x = -2\) into the original equation:

\[
\sqrt{2(-2) + 20} = -2 + 6
\]

\[
\sqrt{-4 + 20} = 4
\]

\[
\sqrt{16} = 4
\]

Since this is true, \(x = -2\) is a valid solution.

#### Checking \(x = -8\):

Substitute \(x = -8\) into the original equation:

\[
\sqrt{2(-8) + 20} = -8 + 6
\]

\[
\sqrt{-16 + 20} = -2
\]

\[
\sqrt{4} = -2
\]

This is false because the square root of 4 is positive (2), not negative. Therefore, \(x = -8\) is not a valid solution.

### Final Solution:

The only valid solution is \(x = -2\).

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How do you solve a radical equation with multiple terms inside the square root?
2. What happens if a quadratic equation has complex solutions?
3. How can you simplify the process of solving radical equations?
4. Can a square root ever yield a negative number as a result?
5. What is the significance of checking solutions in the original equation?

**Tip:** Always remember to check for extraneous solutions when solving radical equations by plugging the answers back into the original equation.

Solve the given radical equation. Check all proposed solutions. StartRoot 2 x plus 20 EndRoot equals x plus 6.

Step-by-step guide on solving the radical equation √(2x + 20) = x + 6, leading to a quadratic equation. Includes methods for checking solutions to avoid extraneous results. Ideal for students in Grades 9-11.

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