Question 534075
Let the smaller number be x
Then the larger number be {{{x^2/8}}}


{{{(x^2/8)^2-x^2=180}}}
{{{(x^(2(2))/8^2)-x^2=180}}}
{{{x^4/64-x^2=180}}}
{{{64(x^4/64)-64(x^2)=64(180)}}}
{{{x^4-64x^2=11520}}}
{{{x^4-64x^2-11520=0}}}
{{{(x^2+80)(x^2-144)=0}}}
{{{(x^2+80)(x-12)(x+12)=0}}}
1. x^2+80=0
2. x-12=0  ***Set each of the factors of the equation equal to 0***
3. x+12=0


1. {{{x^2= -80}}}
{{{sqrt(x^2)}}}=+_{{{sqrt(-80)}}}
{{{x=+-4i{{{sqrt(5)}}}
x=4i{{{sqrt(5)}}}, -4i{{{sqrt(5)}}}


2. x-12=0
x=12


3. x+12=0
x=-12


So x=4i{{{sqrt(5)}}}, -4i{{{sqrt(5)}}}, 12, -12


For the larger number... you calculate yourself just substitute it (I think you are capable of this because you have already started calculating something so complicated as this.


Note: +_ stands for Positive/Negative sign