Find the value of : $\sqrt{8 + 2\sqrt{8 + 2\sqrt{8 + \dots}}}$

Q: Find the value of : $\sqrt{8 + 2\sqrt{8 + 2\sqrt{8 + \dots}}}$

To solve the expression $\sqrt{8 + 2\sqrt{8 + 2\sqrt{8 + \dots}}}$, we need to find a value $x$ such that: \[ x = \sqrt{8 + 2x} \] Squaring both sides, we get: \[ x^2 = 8 + 2x \] Rearranging the equation gives: \[ x^2 - 2x - 8 = 0 \] This is a quadratic equation, which can be solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, $a = 1$, $b = -2$, and $c = -8$. Plugging these values into the formula, we get: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 + 32}}{2} \] \[ x = \frac{2 \pm \sqrt{36}}{2} \] \[ x = \frac{2 \pm 6}{2} \] This gives us two potential solutions: 1. $ x = \frac{2 + 6}{2} = 4 $ 2. $ x = \frac{2 - 6}{2} = -2 $ Since $x$ represents a length (as it is under a square root), it must be non-negative. Therefore, the only valid solution is $x = 4$. Now, let's examine why the other options are incorrect: 1. **Option 1 (8):** If $x = 8$, substituting back into the equation $x = \sqrt{8 + 2x}$ gives $8 = \sqrt{8 + 16} = \sqrt{24}$, which is not true since $\sqrt{24} \neq 8$. 2. **Option 2 (10):** If $x = 10$, substituting back gives $10 = \sqrt{8 + 20} = \sqrt{28}$, which is not true since $\sqrt{28} \neq 10$. 3. **Option 3 (11):** If $x = 11$, substituting back gives $11 = \sqrt{8 + 22} = \sqrt{30}$, which is not true since $\sqrt{30} \neq 11$. Thus, the only correct and valid solution is Option 4, which is $x = 4$.

Question

Accepted Answer

To solve the expression $\sqrt{8 + 2\sqrt{8 + 2\sqrt{8 + \dots}}}$, we need to find a value $x$ such that:

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

Squaring both sides, we get:

$$ x^2 = 8 + 2x $$

Rearranging the equation gives:

$$ x^2 - 2x - 8 = 0 $$

This is a quadratic equation, which can be solved using the quadratic formula:

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

Here, $a = 1$, $b = -2$, and $c = -8$. Plugging these values into the formula, we get:

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

$$ x = \frac{2 \pm \sqrt{4 + 32}}{2} $$

$$ x = \frac{2 \pm \sqrt{36}}{2} $$

$$ x = \frac{2 \pm 6}{2} $$

This gives us two potential solutions:

1. $ x = \frac{2 + 6}{2} = 4 $
2. $ x = \frac{2 - 6}{2} = -2 $

Since $x$ represents a length (as it is under a square root), it must be non-negative. Therefore, the only valid solution is $x = 4$.

Now, let's examine why the other options are incorrect:

1. **Option 1 (8):** If $x = 8$, substituting back into the equation $x = \sqrt{8 + 2x}$ gives $8 = \sqrt{8 + 16} = \sqrt{24}$, which is not true since $\sqrt{24} \neq 8$.

2. **Option 2 (10):** If $x = 10$, substituting back gives $10 = \sqrt{8 + 20} = \sqrt{28}$, which is not true since $\sqrt{28} \neq 10$.

3. **Option 3 (11):** If $x = 11$, substituting back gives $11 = \sqrt{8 + 22} = \sqrt{30}$, which is not true since $\sqrt{30} \neq 11$.

Thus, the only correct and valid solution is Option 4, which is $x = 4$.