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find two consecutive positive integers sum of whose squares is 365

Asked by Ciara(student) , on 1/11/12

Answers

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Let The Numbers Be x And x+1

Squares x2 And (x+1)2

Sum = 365

Hence

x2+(x+1)2 = 365

x 2 +x2+1+2x = 365

2x2+2x+1 = 365

2x2+2x-364 = 0

x2+x-182 = 0

Use Quadratic Formula

-b+Square Root (b2-4ac) / 2a    And -b-Square Root (b2-4ac) / 2a

-1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2

-1+27/2 And -1-27/2

13 And -14

As Positive Integers So 13

Other Integer = 13+1 = 14

Posted by Prakhar Bindal(student)on 1/11/12

BEST ANSWER Certified by MeritNation Expert  
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Let the consecutive positive integers be x and x + 1.

Either x + 14 = 0 or x − 13 = 0, i.e., x = 14 or x = 13

Since the integers are positive, x can only be 13.

x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Posted by Ishan Goyal(student)on 1/11/12

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