Question

$\sqrt{43-12\sqrt{7}}-\frac{2}{\sqrt{16+6\sqrt{7}}}=?$

• A. $2\sqrt{7}-1$
• B. $-2\sqrt{7}+1$
• C. $-3$
• D. $3$
• E. $Noneofthese$

Answer 1

The correct option is D $3$

Here, it can also be written as
$\sqrt{(6{)}^2+(\sqrt{7}{)}^2-2.6\sqrt{7}}-\frac{2}{\sqrt{(\sqrt{7}{)}^2}+(3{)}^2+2\sqrt{7.3}}$
Apply the formula
$a^2+b^2-2ab=(a-b{)}^{2}And{a}^2+b^2+2ab=(a+b{)}^{2}So,\sqrt{(6-\sqrt{7}{)}^2}-\frac{2}{\sqrt{(3+\sqrt{7}{)}^2}}=\sqrt{(6-\sqrt{7})}=\frac{2}{3+\sqrt{7}}=(6-\sqrt{7})-\frac{2}{3+\sqrt{7}}=(6-\sqrt{7})-\frac{2}{(3+\sqrt{7})(3-\sqrt{7})}=(6-\sqrt{7})-\frac{2(3-\sqrt{7})}{(9-7)}$
$[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$
$=(6-\sqrt{7})-(3-\sqrt{7})=6-\sqrt{7}-3+\sqrt{7}=3$