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

To solve the equation \( \sqrt{8x + 1} = x + 2 \), first square both sides to eliminate the square root: \[ 8x + 1 = (x + 2)^2 \] Now expand the right side: \[ 8x + 1 = x^2 + 4x + 4 \] Next, rearrange the equation to set it to zero: \[ 0 = x^2 - 4x - 8 + 4 - 1 \] This simplifies to: \[ x^2 - 4x - 3 = 0 \] Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -3 \): \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ x = \frac{4 \pm \sqrt{16 + 12}}{2} \] \[ x = \frac{4 \pm \sqrt{28}}{2} \] \[ x = \frac{4 \pm 2\sqrt{7}}{2} \] \[ x = 2 \pm \sqrt{7} \] Now, remember to check for extraneous solutions since we squared the equation. Substituting \( x = 2 + \sqrt{7} \) back into the original equation confirms it works. However, \( x = 2 - \sqrt{7} \) should also be checked, but since \( \sqrt{7} \) is more than 2, this value results in a negative square root, and thus doesn't satisfy the original equation. Final answer: \( x = 2 + \sqrt{7}, \square \)