Q7. 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

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

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

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 = \(y = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

=>\(y = {{8 ± \sqrt{(-8)^2 - 4(1)(-180)}} \over {(2)(1)}}\)

=>\(y = {{8 ± \sqrt{64 + 720}} \over {2}} = {{60 ± \sqrt{784}} \over {2}}\)

=>\(y = {{8 + 28} \over {2}} , {{8 - 28} \over {2}}\)

=>\(y = 18,-10\)

Using equation (2) to find smaller number:

\(x^2 = 8y\)

=>\(x^2 = 8y = (8)(18) = 144\)

=>\( x = ±12\)

And, = \(x^2 = 8y = (8)(-10) = -80\) {No real solution for x}

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