Question

Which of the following surds is greatest? ³√2 , √3 , ⁶√7 , ⁴√5 *
4 points
³√2
√3
⁴√5
⁶√7

Answer 1

Answer:

a

Step-by-step explanation:

Answer 2

$\large\underline{\sf{Solution-}}$

Given surds are

$\red{\rm :\longmapsto\: \sqrt[3]{2}, \: \sqrt{3}, \: \sqrt[6]{7}, \: \sqrt[4]{5}}$

To find which surd is greater, Let we first convert them in to same powers.

Now, let first find the LCM of ( 3, 2, 6, 4 ) = 12

We know,

$\boxed{ \bf{ \: \sqrt[x]{y} \: = \: \sqrt[xz]{ {y}^{z} }}}$

So,

$\rm :\longmapsto\: \sqrt[3]{2} = \sqrt[3 \times 4]{ {2}^{4} } = \sqrt[12]{16}$

$\rm :\longmapsto\: \sqrt[2]{3} = \sqrt[3 \times 6]{ {3}^{6} } = \sqrt[12]{729}$

$\rm :\longmapsto\: \sqrt[6]{7} = \sqrt[2 \times 6]{ {7}^{2} } = \sqrt[12]{49}$

$\rm :\longmapsto\: \sqrt[4]{5} = \sqrt[4 \times 3]{ {5}^{3} } = \sqrt[12]{125}$

So, from these, we concluded that,

$\rm :\longmapsto\: \sqrt[12]{729} > \sqrt[12]{125} > \sqrt[12]{49} > \sqrt[12]{16}$

$\bf\implies \: \sqrt[2]{3} > \sqrt[4]{5} > \sqrt[6]{7} > \sqrt[3]{2}$

$\bf\implies \: \sqrt{3} \: is \: greatest$

• Hence, option (2) is correct