Question

Find two consecutive positive odd integers, sum of whose squares is 290
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Answer 1

Answer:

Step-by-step explanation:

Answer is 13 and 14

Answer 2

Answer:

Let the first positive odd integer be x

another odd integer be (x+2)

square:

x^2(x+2)^2=290

x^2×x^2+4+4x=290

2x^2+4x+4=290

2x^2+4x=286=0

2x^2+4x-286=0

divide by 2

x^2+2x-143=0

x^2+13x-11x-143=0

x(x+13)-11(x+13)

(x-11)=0 (x+13)=0

x=11. x=-13

Positive number should be taken

So, the positive odd integer

Another positive odd integer (x+2)=11+2 =13

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