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# Surds Rationalising the denominator

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Apr 11, 2018

#1
+7616
+2

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     $$b=4$$  .

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Apr 11, 2018
edited by hectictar  Apr 11, 2018

#1
+7616
+2

Given that   $$\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2$$   where  $$a$$  and  $$b$$  are integers, find the values of  $$a$$  and  $$b$$ .

----------------

$$\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2$$

So     $$a=-3$$     and     $$b=4$$  .

hectictar Apr 11, 2018
edited by hectictar  Apr 11, 2018