Question

find the number whose square root is twice its cube root.

Answer 1

See the attached file for solution. Thanks

Answer 2

Answer:

The number which has its square root will be equal to twice of its cube root will be 64.

Solution:

Let the number be taken as x.

Given the expression will be,

$\begin{array} { l } { \sqrt { x } = 2 \times \sqrt [ 3 ] { x } } \\\\ { x ^ { \frac { 1 } { 2 } } = 2 \times x ^ { \frac { 1 } { 3 } } } \\\\ { \quad \frac { x ^ { \frac { 1 } { 2 } } } { x ^ { \frac { 1 } { 3 } } } = 2 } \\\\ { x \left( \frac { 1 } { 2 } - \frac { 1 } { 3 } \right) = 2 } \end{array}$

$\begin{array} { c } { x ^ { \frac { 3 - 2 } { 6 } } = 2 } \\\\ { x ^ { \frac { 1 } { 6 } } = 2 } \\\\ { \left( x ^ { \frac { 1 } { 6 } } \right) ^ { 6 } = 2 ^ { 6 } } \\\\ { x = 2 ^ { 6 } } \\\\ { x = 64 } \end{array}$

Therefore, the number which has its square root will be equal to twice of its cube root will be 64.