22. Given that \[ \frac{10-\sqrt{32}}{\sqrt{2}}=a+b \sqrt{2} \] where a and b are integer. Find the values of \( a \) and \( b \). \[ \begin{array}{l} a=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ b=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{array} \] (4)

To solve the equation \[ \frac{10-\sqrt{32}}{\sqrt{2}}=a+b \sqrt{2}, \] we first simplify the left-hand side. We know that \(\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}\). Therefore, we can rewrite our equation as: \[ \frac{10 - 4\sqrt{2}}{\sqrt{2}}. \] Now, let's separate the fraction: \[ \frac{10}{\sqrt{2}} - \frac{4\sqrt{2}}{\sqrt{2}} = \frac{10}{\sqrt{2}} - 4. \] Next, we rationalize \(\frac{10}{\sqrt{2}}\): \[ \frac{10}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}. \] Combining everything, we get: \[ 5\sqrt{2} - 4 = -4 + 5\sqrt{2}. \] In this form, we can identify \(a\) and \(b\): \[ a = -4, \quad b = 5. \] Thus, \[ \begin{array}{l} a=-4 \\ b=5 \end{array} \]