The difference of squares of two …

Let the larger number be x,
Then, (smaller number)² = 8(larger number)
= 8x {tex}\Rightarrow{/tex} smaller number {tex}= \sqrt { 8 x }{/tex}
According to the question, x² - 8x = 180
{tex}\Rightarrow x ^ { 2 } - 8 x - 180 = 0{/tex}
Using the quadratic formula, {tex}= \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}
we get {tex}= \frac { - ( - 8 ) \pm \sqrt { ( - 8 ) ^ { 2 } - 4 ( 1 ) ( - 180 ) } } { 2 ( 1 ) }{/tex}
{tex}= \frac { 8 \pm \sqrt { 64 + 720 } } { 2 } = \frac { 8 \pm \sqrt { 784 } } { 2 } = \frac { 8 \pm 28 } { 2 }{/tex}
{tex}= \frac { 8 + 28 } { 2 } , \frac { 8 - 28 } { 2 } \Rightarrow x = 18 , - 10{/tex}
x = -10 is inadmissible
Then smaller number {tex}= \sqrt { 8 ( - 10 ) } = \sqrt { - 80 }{/tex}
which does not exist.
{tex}\therefore x = 18 \therefore \sqrt { 8 x } = \sqrt { 8 \times 18 } = \sqrt { 144 } = \pm 12{/tex}
Hence, the two numbers are 18, 12 or 18, -12.