The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let smaller number = x and let larger number = y

According to condition:
=>\(y^2 - x^2 = 180\)… (1)

Also, we are given that square of smaller number is 8 times the larger number.
=>\(x^2 = 8y\) … (2)

Putting equation (2) in (1), we get

\(y^2 - 8y = 180\)
\(\Rightarrow \) \(y^2 - 8y - 180 = 0\)

Comparing equation \(y^2 - 8y - 180 = 0\) with general form \(ax^2 + bx + c = 0\),

We get a = 1, b = -8 and c = -180

Using quadratic formula = \(y = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
\(\Rightarrow \) \(y = {{8 ± \sqrt{(-8)^2 - 4(1)(-180)}} \over {(2)(1)}}\)
\(\Rightarrow \) \(y = {{8 ± \sqrt{64 + 720}} \over {2}} = {{60 ± \sqrt{784}} \over {2}}\)
(\Rightarrow \) \(y = {{8 + 28} \over {2}} , {{8 - 28} \over {2}}\)
(\Rightarrow \) \(y = 18,-10\)

Using equation (2) to find smaller number:
\(x^2 = 8y\)
(\Rightarrow \) \(x^2 = 8y = (8)(18) = 144\)
(\Rightarrow \) \( x = ±12\)
And, = \(x^2 = 8y = (8)(-10) = -80\) {No real solution for x}

Therefore, two numbers are (12, 18) or (-12, 18)