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Question

# Find rational numbers a and b such that (i) $\frac{\sqrt{2}-1}{\sqrt{2}+1}=a+b\sqrt{2}$ (ii) $\frac{2-\sqrt{5}}{2+\sqrt{5}}=a\sqrt{5}+b$ (iii) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}$ (iv) $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$

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Solution

## (i) $\frac{\sqrt{2}-1}{\sqrt{2}+1}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{2}-1\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}$ $=\frac{2+1-2\sqrt{2}}{2-1}\phantom{\rule{0ex}{0ex}}=3-2\sqrt{2}$ $\therefore \frac{\sqrt{2}-1}{\sqrt{2}+1}=3+\left(-2\right)\sqrt{2}=a+b\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒a=3,b=-2$ (ii) $\frac{2-\sqrt{5}}{2+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{5}}{2+\sqrt{5}}×\frac{2-\sqrt{5}}{2-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2-\sqrt{5}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$ $=\frac{4+5-4\sqrt{5}}{4-5}\phantom{\rule{0ex}{0ex}}=\frac{9-4\sqrt{5}}{-1}\phantom{\rule{0ex}{0ex}}=-9+4\sqrt{5}$ $\therefore \frac{2-\sqrt{5}}{2+\sqrt{5}}=4\sqrt{5}+\left(-9\right)=a\sqrt{5}+b\phantom{\rule{0ex}{0ex}}⇒a=4,b=-9$ (iii) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}×\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$ $=\frac{3+2+2×\sqrt{3}×\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=5+2\sqrt{6}$ $\therefore \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2\sqrt{6}=a+b\sqrt{6}\phantom{\rule{0ex}{0ex}}⇒a=5,b=2$ (iv) $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}$ $=\frac{11-6\sqrt{3}}{49-48}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$ $\therefore \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒a=11,b=-6$

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