Question

$\frac{5+\sqrt{2}}{5-\sqrt{2}}=m+n\sqrt{2}$

Find the value of m and n

• A. We define for any positive integer 'n' and all positive real values of 'a'
• B. $m=3andn=\sqrt{2}$
• C. $m=9andn=\frac{10}{3}$
• D. $m=27andn=10$

Answer 1

The correct option is C

$m=9andn=\frac{10}{3}$

Given that

$\frac{5+\sqrt{2}}{5-\sqrt{2}}=m+n\sqrt{2}$
R.F of the denominator = 5 + $\sqrt{2}$

Multiplying both numerator and denominator by Rationalizing factor, we get

$\frac{5+\sqrt{2}}{5-\sqrt{2}}$ $\times$ $\frac{5+\sqrt{2}}{5+\sqrt{2}}$ = $m+n\sqrt{2}$

$\frac{(5+\sqrt{2}{)}^2}{(5{)}^2-\sqrt{\left(2\right)}{)}^2}$ = $m+n\sqrt{2}$ [( $a^2-b^2=(a+b)(a-b)$)]

$\frac{25+10\sqrt{2}+2}{3}$ = $m+n\sqrt{2}$ [$(a+b{)}^2=a^2+2ab+b^2$]

$\frac{27+10\sqrt{2}}{3}$ = $m+n\sqrt{2}$

$\frac{27}{3}+\frac{10\sqrt{2}}{3}$ = $m+n\sqrt{2}$

Comparing the coefficient , we have

$m=9andn=\frac{10}{3}$