Question
In the following determine rational numbers a and b:
$\frac{4+\sqrt2}{2+\sqrt2}=\text{a}-\sqrt{\text{b}}$
$\frac{4+\sqrt2}{2+\sqrt2}=\text{a}-\sqrt{\text{b}}$
✓
Answer
We know that rationalization factor for $2+\sqrt2$ is $2-\sqrt2.$ We will multiply numerator and denominator of the given expression $\frac{4+\sqrt2}{2+\sqrt2}$ by $2-\sqrt2,$ to get
$\frac{4+\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{4\times2-4\times\sqrt2+2\times\sqrt2-\big(\sqrt2\big)^2}{(2)^2-\big(\sqrt2\big)^2}$
$=\frac{8-4\sqrt2+2\sqrt2-2}{4-2}$
$=\frac{6-2\sqrt2}{2}$
$=3-\sqrt2$
On equating rational and irrational terms, we get
$\text{a}-\sqrt{\text{b}}=3-\sqrt2$
Hence, we get a = 3, b = 2.
$\frac{4+\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{4\times2-4\times\sqrt2+2\times\sqrt2-\big(\sqrt2\big)^2}{(2)^2-\big(\sqrt2\big)^2}$
$=\frac{8-4\sqrt2+2\sqrt2-2}{4-2}$
$=\frac{6-2\sqrt2}{2}$
$=3-\sqrt2$
On equating rational and irrational terms, we get
$\text{a}-\sqrt{\text{b}}=3-\sqrt2$
Hence, we get a = 3, b = 2.
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