Question

In the following expression, find the values of $p$ and $q$

$\frac{\sqrt{2}+2}{\sqrt{2}-1}=p+q\sqrt{2}$

• A. $p=0,q=1$
• B. $p=1,q=0$
• C. $p=1,q=1$
• D. $p=4,q=3$

Answer 1

The correct option is D $p=4,q=3$

$\frac{\sqrt{2}+2}{\sqrt{2}-1}=p+q\sqrt{2}$

$\text{RHS}=p+q\sqrt{2}$

$\text{LHS}=\frac{\sqrt{2}+2}{\sqrt{2}-1}$

Rationalizing the denominator of $\text{LHS}$, we get

$\text{LHS}=\frac{\sqrt{2}+2}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

Note: $(a+b)(a-b)=a^2-b^2$

$\text{LHS}=\frac{(\sqrt{2}+2)(\sqrt{2}+1)}{(\sqrt{2}{)}^2-1^2}$

$=\frac{2+3\sqrt{2}+2}{2-1}$

$=\frac{4+3\sqrt{2}}{1}$

$=4+3\sqrt{2}$

Since, $\text{LHS = RHS}$
Hence,
$p+q\sqrt{2}=4+3\sqrt{2}$

$\therefore$ $p=4,q=3$