Let the larger number be \(x\), Then the other number will be \((16 - x)\)
According to the question, \(2 \times x^2 = (16 - x)^2 + 164\) Or, \(2x^2 = 16^2 - 32x + x^2 + 164\) Or, \(2x^2 - x^2 + 32x - 256 - 164 = 0\) Or, \(x^2 + 32x - 420 = 0\) Or, \(x^2 + (42 - 10)x - 42 = 0\) Or, \(x^2 + 42x - 10x - 42 = 0\) Or, \(x(x + 42) - 10(x + 42) = 0\) Or, \((x + 42)(x - 10) = 0\)
∴ Either \((x + 42) = 0\) or \((x - 10) = 0\) When \((x + 42) = 0\), then \(x = -42\) [Negative, not acceptable] When \((x - 10) = 0\), then \(x = 10\)
∴ The larger number is 10 and the smaller number is \(16 - 10 = 6\).