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# 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)