Question

Prove that there is a unique pair of consecutive odd positive intergers such that sum of

their squares is 290, find them​

Answer 1

Answer:

this is ur answer ☺

Step-by-step explanation:

Let x an odd positive integer

Then, according to question

x

2

+(x+2)

2

=290

2x

2

+4x−286=0

x

2

+2x−143=0

x

2

+13x−11x−143=0

(x+13)(x−11)=0

x=11 as x is positive

Hence required integers are 11 and 13

Answer 2

Answer:

ANSWER

Let x an odd positive integer

Then, according to question

x2+(x+2) 2 =290

2x2 +4x−286=0

x2 +2x−143=0

x2 +13x−11x−143=0

(x+13)(x−11)=0

x=11 as x is positive

Hence required integers are 11 and 13.

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