Applying Difference of Perfect Squares

Algebra Level pending

Given that x 2 − y 2 = ( x − y ) ( x + y ) x^2-y^2=(x-y)(x+y) , what is the value of 3 1 2 − 1 9 2 31^2-19^2 ?


The answer is 600.

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2 solutions

Mahdi Raza
Aug 20, 2020

3 1 2 − 1 9 2 = ( 31 + 19 ) ( 31 − 19 ) = ( 12 ) ( 50 ) = 600 31^2 - 19^2 = (31+19)(31-19) = (12)(50) = \boxed{600}

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Brilliant Mathematics Staff
Aug 1, 2020

In this case, x = 31 x = 31 and y = 19 y = 19 . Hence, 3 1 2 − 1 9 2 = x 2 − y 2 = ( x − y ) ( x + y ) = ( 31 − 19 ) ( 31 + 19 ) = ( 12 ) ( 50 ) = 600 . 31^2 - 19^2 = x^2 - y^2 = (x - y)(x + y) = (31 - 19)(31 + 19) = (12)(50) = \boxed{600}.

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