Solution:

Let us say, the larger and smaller number be x and y, respectively.

As per the question given,

${{x}^{2~}}~{{y}^{2}}~=\text{ }180\text{ }and~{{y}^{2}}~=\text{ }8x$

$\Rightarrow ~{{x}^{2~}}\text{ }8x~=\text{ }180$

$\Rightarrow ~{{x}^{2~}}~8x~\text{ }180\text{ }=\text{ }0$

$\Rightarrow ~{{x}^{2~}}\text{ }18x~+\text{ }10x~\text{ }180\text{ }=\text{ }0$

$\Rightarrow ~x\left( x~\text{ }18 \right)~+10\left( x~\text{ }18 \right)\text{ }=\text{ }0$

$\Rightarrow \left( x~\text{ }18 \right)\left( x~+\text{ }10 \right)\text{ }=\text{ }0$

$\Rightarrow ~x~=\text{ }18,\text{ }-10$

However, the larger number cannot be considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

$x~=\text{ }18$

$\therefore ~{{y}^{2}}~=\text{ }8x\text{ }=\text{ }8\text{ }\times \text{ }18\text{ }=\text{ }144$

$\Rightarrow ~y~=\text{ }\pm \surd 144~=\text{ }\pm 12$

∴ Smaller number $=\text{ }\pm 12$

Therefore, the numbers are 18 and 12 or 18 and -12.