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Find two consecutive positive integers such that the sum of their squares is 421 .

1 Answer

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We assume the two numbers are x and (x+1)
so x^2+(x+1)^2=421
x^2+(x^2+2x+1)=421
2x^2+2x-420=0
According to the formula of quadratic
x=14 or-15
cuz we know the two numbers are integers
so x=14
therefore the other number is 15
To make sure that's correct
14^2+15^2=421

Hope that helps you!!


User Perh
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