Question

Compare the following pair of surds.

(i) $7\sqrt{2},5\sqrt{3}$ (ii) $\sqrt{247},\sqrt{274}$ (iii) $2\sqrt{7},\sqrt{28}$ (iv) $5\sqrt{5},7\sqrt{2}$ (v) $4\sqrt{42},9\sqrt{2}$ (vi) $5\sqrt{3},9$ (vii) $7,2\sqrt{5}$

Answer 1

(i) $7\sqrt{2},5\sqrt{3}$

Since, $7\sqrt{2}=\sqrt{49\times 2}=\sqrt{98}\mathrm{and}5\sqrt{3}=\sqrt{25\times 3}=\sqrt{75}$

So, $7\sqrt{2}>5\sqrt{3}$

(ii) $\sqrt{247},\sqrt{274}$

$\sqrt{247}<\sqrt{274}$

(iii) $2\sqrt{7},\sqrt{28}$

Since, $2\sqrt{7}=\sqrt{4\times 7}=\sqrt{28}$

So, $2\sqrt{7}=\sqrt{28}$

(iv) $5\sqrt{5},7\sqrt{2}$

Since, $5\sqrt{5}=\sqrt{25\times 5}=\sqrt{125}\mathrm{and}7\sqrt{2}=\sqrt{49\times 2}=\sqrt{98}$

So, $5\sqrt{5}>7\sqrt{2}$

(v) $4\sqrt{42},9\sqrt{2}$

Since, $4\sqrt{42}=\sqrt{16\times 42}=\sqrt{672}\mathrm{and}9\sqrt{2}=\sqrt{81\times 2}=\sqrt{162}$

So, $4\sqrt{42}>9\sqrt{2}$

(vi) $5\sqrt{3},9$

Since, $5\sqrt{3}=\sqrt{25\times 3}=\sqrt{75}\mathrm{and}9=\sqrt{81}$

So, $5\sqrt{3}<9$

(vii) $7,2\sqrt{5}$

Since, $7=\sqrt{49}\mathrm{and}2\sqrt{5}=\sqrt{4\times 5}=\sqrt{20}$

So, $7>2\sqrt{5}$