After rationalising the denominator
Question:

After rationalising the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}}$, we get the denominator as

(a) 13

(b) 19

(c) 5

(d) 35

 

Solution:

$\frac{7}{3 \sqrt{3}-2 \sqrt{2}}=\frac{7}{3 \sqrt{3}-2 \sqrt{2}} \cdot \frac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}}$        [multiplying numerator and denominator by $3 \sqrt{3}+2 \sqrt{2}$ ]

$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}$

[using identity $(a-b)(a+b)=a^{2}-b^{2}$ ]

$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{27-8}$                  $\left[\because\right.$ fraction $\left.=\frac{\text { numerator }}{\text { denominator }}\right]$

$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{19}$

Hence, after rationalising the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}}$, we get the denominator as 19 .

 

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