the sum of the squares of two consecutive odd positive integers is 394. Find them
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Let the two numbers are x and(x+2)
According to the question (x) ^2+(x+2)^2=394
=>x^2+x^2+4x+4=394
=>2(x)^2+4x+4-394=0
=>2(x)^2+4x-390=0
=>2(x)^2+30x-26x-390=0
=>2x(x+15)-26(x+15)=0
=>(x+15)(2x-26)=0
=>x+15=0 or 2x-26=0
=>x= -15 orx=26/2=13
if x= -15,then x+2= -13
or
ifx=13,thenx+2=15
According to the question (x) ^2+(x+2)^2=394
=>x^2+x^2+4x+4=394
=>2(x)^2+4x+4-394=0
=>2(x)^2+4x-390=0
=>2(x)^2+30x-26x-390=0
=>2x(x+15)-26(x+15)=0
=>(x+15)(2x-26)=0
=>x+15=0 or 2x-26=0
=>x= -15 orx=26/2=13
if x= -15,then x+2= -13
or
ifx=13,thenx+2=15
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