Question

Which of the below surds are arranged in decending order?

• A. $11\sqrt{2}$>$7\sqrt{7}$<$8\sqrt{6}$<$10\sqrt{3}$
• B. $8\sqrt{6}$>$7\sqrt{7}$>$10\sqrt{3}$>$11\sqrt{2}$
• C. $8\sqrt{6}$<$7\sqrt{7}$<$10\sqrt{3}$>$11\sqrt{2}$
• D. $11\sqrt{2}$<$7\sqrt{7}$<$8\sqrt{6}$<$10\sqrt{3}$

Answer 1

The correct option is B $8\sqrt{6}$>$7\sqrt{7}$>$10\sqrt{3}$>$11\sqrt{2}$

If x>y
Then, $x^2>y^2$
So, squaring all the surds
$(11\sqrt{2}{)}^2=121\times 2=242$
$(7\sqrt{7}{)}^2=49\times 7=343$
$(8\sqrt{6}{)}^2=384$
$(10\sqrt{3}{)}^2=100\times 3=300$
So, the correct decending order is given by,
384>343>300>242
That is,
$8\sqrt{6}$>$7\sqrt{7}$>$10\sqrt{3}$>$11\sqrt{2}$