+412 https://vle.mathswatch.co.uk/images/questions/question1918.png
+7348 Given that \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\) where \(a\) and \(b\) are integers, find the values of \(a\) and \(b\) .
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\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)
So \(a=-3\) and \(b=4\) .
.+7348 Given that \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\) where \(a\) and \(b\) are integers, find the values of \(a\) and \(b\) .
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\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)
So \(a=-3\) and \(b=4\) .
