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

Question

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

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Piper 2 months 2021-07-27T22:45:37+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-07-27T22:46:43+00:00

    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=+13x11x

    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 : x11=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 .

    0
    2021-07-27T22:47:19+00:00

    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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