$$ (x+2)^{\frac{1}{2}}=x $$ What is the solution to the given equation?

Question

UpStudy AI · Accepted Answer

To solve the equation:

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

**Step 1: Determine the Domain**

The square root function requires the argument to be non-negative:

$$
x + 2 \geq 0 \implies x \geq -2
$$

Additionally, since $$\sqrt{x + 2} = x$$ and the square root is always non-negative, $$x$$ must also satisfy:

$$
x \geq 0
$$

So, the domain of the equation is $$x \geq 0$$.

**Step 2: Square Both Sides**

To eliminate the square root, square both sides of the equation:

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

**Step 3: Rearrange the Equation**

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

$$
x^2 - x - 2 = 0
$$

**Step 4: Solve the Quadratic Equation**

Use the quadratic formula:

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

For the equation $$x^2 - x - 2 = 0$$, $$a = 1$$, $$b = -1$$, and $$c = -2$$. Plugging these in:

$$
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}
$$

This gives two potential solutions:

$$
x = \frac{1 + 3}{2} = 2 \quad \text{and} \quad x = \frac{1 - 3}{2} = -1
$$

**Step 5: Verify the Solutions**

Recall the domain $$x \geq 0$$:

- $$x = 2$$ is valid.
- $$x = -1$$ is **not** valid since it doesn't satisfy $$x \geq 0$$.

**Conclusion**

The only real solution to the equation is:

$$
\boxed{2}
$$
The solution to the equation $$ \sqrt{x + 2} = x $$ is $$ x = 2 $$.

\( (x+2)^{\frac{1}{2}}=x \) What is the solution to the given equation?