given That \( 356_{x}=B C_{16} \) find the value of \( x \).

To solve for \( x \) given the equation \( 356_{x} = BC_{16} \), we first need to convert both sides into base 10. Starting with \( BC_{16} \): - In base 16, \( B = 11 \) and \( C = 12 \). - Therefore, \( BC_{16} = 11 \cdot 16^1 + 12 \cdot 16^0 = 11 \cdot 16 + 12 \cdot 1 = 176 + 12 = 188_{10} \). Next, we convert \( 356_{x} \) into base 10: - \( 356_{x} = 3 \cdot x^2 + 5 \cdot x^1 + 6 \cdot x^0 = 3x^2 + 5x + 6 \). Setting the two expressions equal gives us: \[ 3x^2 + 5x + 6 = 188 \] Now we simplify: \[ 3x^2 + 5x + 6 - 188 = 0 \] \[ 3x^2 + 5x - 182 = 0 \] Next, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 5 \), and \( c = -182 \): - Calculate the discriminant: \( b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot (-182) = 25 + 2184 = 2209 \). - Now we can find \( x \): \[ x = \frac{-5 \pm \sqrt{2209}}{2 \cdot 3} \] Since \( \sqrt{2209} = 47 \): \[ x = \frac{-5 \pm 47}{6} \] Calculating the two possible values: 1. \( x = \frac{42}{6} = 7 \) (valid since \( x \) must be greater than the largest digit in base \( x \), which is 6). 2. \( x = \frac{-52}{6} \) (invalid because it's negative). Thus, the solution is: \[ x = 7 \]