Question
In the following determine rational numbers a and b:$\frac{3+\sqrt2}{3-\sqrt2}=\text{a}+\text{b}\sqrt{2}$
✓
Answer
We know that rationalization factor for $3-\sqrt2$ is $3+\sqrt2.$ We will multiply numerator and denominator of the given expression $\frac{3+\sqrt2}{3-\sqrt2}$ by $3+\sqrt2,$ to get$\frac{3+\sqrt2}{3-\sqrt2}\times\frac{3+\sqrt2}{3+\sqrt2}=\frac{(3)^2+\big(\sqrt2\big)^2+2\times3\times\sqrt2}{(3)^2-\big(\sqrt2\big)^2}$
$=\frac{9+2+6\sqrt2}{9-2}$
$=\frac{11+6\sqrt2}{7}$
$=\frac{11}{7}+\frac{6}{7}\sqrt2$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt2=\frac{11}{6}+\frac{6}{7}\sqrt2$
Hence, we get $\text{a}=\frac{11}{7},\ \text{b}=\frac{6}{7}.$
$=\frac{9+2+6\sqrt2}{9-2}$
$=\frac{11+6\sqrt2}{7}$
$=\frac{11}{7}+\frac{6}{7}\sqrt2$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt2=\frac{11}{6}+\frac{6}{7}\sqrt2$
Hence, we get $\text{a}=\frac{11}{7},\ \text{b}=\frac{6}{7}.$
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