Solve the equation. $$ \sqrt{8 \mathrm{x}+1}=\mathrm{x}+2 $$ $$ \mathrm{x}=[?], \square $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
\sqrt{8x + 1} = x + 2
$$

**Step 1: Determine the Domain**

1. The expression inside the square root must be non-negative:
   $$
   8x + 1 \geq 0 \implies x \geq -\frac{1}{8}
   $$
2. The right side of the equation must also be non-negative:
   $$
   x + 2 \geq 0 \implies x \geq -2
   $$
   
Combining both conditions, the domain is $$ x \geq -\frac{1}{8} $$.

**Step 2: Square Both Sides**

$$
\sqrt{8x + 1} = x + 2 \implies (\sqrt{8x + 1})^2 = (x + 2)^2
$$
$$
8x + 1 = x^2 + 4x + 4
$$

**Step 3: Rearrange the Equation**

$$
0 = x^2 + 4x + 4 - 8x - 1
$$
$$
0 = x^2 - 4x + 3
$$

**Step 4: Solve the Quadratic Equation**

$$
x^2 - 4x + 3 = 0
$$
$$
x = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}
$$
$$
x = 3 \quad \text{or} \quad x = 1
$$

**Step 5: Verify the Solutions**

1. **For $$ x = 1 $$:**
   $$
   \sqrt{8(1) + 1} = \sqrt{9} = 3
   $$
   $$
   1 + 2 = 3 \quad \text{(Valid)}
   $$
   
2. **For $$ x = 3 $$:**
   $$
   \sqrt{8(3) + 1} = \sqrt{25} = 5
   $$
   $$
   3 + 2 = 5 \quad \text{(Valid)}
   $$

**Conclusion:**

The solutions to the equation are:

$$
x = 1 \quad \text{and} \quad x = 3
$$

**Final Answer:**

$$
\boxed{\left\{\,1,\;3\,\right\}}
$$
The solutions are $$ x = 1 $$ and $$ x = 3 $$.

Solve the equation.

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