find two consecutive odd positive integers ,sum of whose squares is 290​

2 thoughts on “find two consecutive odd positive integers ,sum of whose squares is 290​”

1. Answer:

   11 and 13

   Step-by-step explanation:

   let the two consecutive odd prime integers be = x and x+2

   the sum of both the numbers are 290

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

   value of x => (x+2)^2=x^2+2x+4

   => x^2+x^2+2x+4=290

   => 2x^2+4x-286=0

   => x^2+2x-143=0

   We need to find two numbers whose sum is 2 and product is -143

   there are two numbers +13 and -11 whose sum is 2 and product is -143

   therefore , 2x=+13x–11x

   by putting this in the quadratic equation ,

   => x^2+2x-143=0

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

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

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

   first number is : x+13=0

   x=> –13

   -13 is a negative odd number and we are asked to find two consecutive positive odd numbers therefore , -13 Isn’t our answer

   second number is : x–11=0

   x=> 11

   we have taken the second odd number as x+2

   x+2=11+2=13

   So , the correct answers are 11 and 13 .

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2. Answer:

   let the 2 numbers be x&(x+2)

   Step-by-step explanation:

   the sum of the squares of numbers=290

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

   x^2+x^2+2^2+2×2×x=290

   2x^2+4+4x=290

   x^2+2x+2=290/2

   x^2+2x=145-2

   x^2+2x-143=0

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

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

   x-11=0 x+13=0

   x=11 x=-13

   the two consecutive numbers are 11 &13.

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