Two integers have a difference of 4. The sum of their squares is 250. Determine the integers.?

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  • 1 month ago

    Let the integers be x, y & x>y then

    x-y=4

    x^2+y^2=250

    =>

    (y+4)^2+y^2=250

    =>

    2y^2+8y+16=250

    =>

    y^2+4y-117=0

    =>

    (y+13)(y-9)=0

    =>

    y=-13 or y=9

    and

    x=-9 or x=13

    correspondingly.

    =>

    answers:

    x=-9 & y=-13

    or

    x=13 & y=9

  • 1 month ago

    the numbers are 13 and 9.

  • 1 month ago

    There are 2 possible answers.

    It could be (13, 9) or (-9, -13)

  • 1 month ago

    9^2=81

    13^2=169

    169+81=250

    9 &13

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  • 1 month ago

    Let the integers be 'm' & 'n' 

    m - n = 4 

    m^2 + n^2 = 250 

    Hence 

    m = 4 + n 

    Substitute 

    (4 + n)^2 + n^2 = 250 

    16 + 8n + n^2 + n^2 = 250 

    2n^2 + 8n - 234 = 0 

    2(n^2 + 4n - 117) = 0 

    Factor 

    (n - 9)(n + 13) = 0 

    Hence n = 9  & m = 13 

  • Dixon
    Lv 7
    1 month ago

    Seeing as 4 << 250, the integers are similar and they will be approx 

    √(250/2) ± 2 

    = 11.1 ± 2

    = 9 , 13

    Check

    9² + 13² = 81 + 169 = 250

  • 1 month ago

    ley x and y are integers

    x - y = 4 eq1

    x^2 + y^2 = 250 eq2

    from eq1 let x your subject.

    x - y = 4

    x = y + 4

    plug in eq1 to eq2

    x^2 + y^2 = 250

    (y + 4)^2 + y^2 = 250

    y^2 + 8y + 16 + y^2 = 250

    2y^2 + 8y + 16 - 250 = 0

    2y^2 + 8y - 234 = 0

    y^2 + 4y -117 = 0

    (y + 13)(y - 9) = 0

    y + 13 = 0, y - 9 = 0

    y = - 13, y = 9

    solving for x from eq1

    when y = - 13

    x = - 13 + 4 = -9

    when y = 9

    x = 9 + 4 = 13

    The integers are: -13, 9, -9, 13.. 

  • 1 month ago

     Two integers have a difference of 4. 

     The sum of their squares is 250. 

     Determine the integers.

     x - y = 4

     x^2 + y^2 = 250

     2y^2 + 8y - 234 = 0

     2(y^2 + 4y - 117) = 0

     y = 9

     The two integers are 13 and 9.

  • 1 month ago

    x-y=4

    x^2 + y^2=250

    y=4+x

    x^2 +(4+ x)^2 =250

    x^2 + x^(2)+8*x+16= 250

    2x^2 + 8x - 234=0

    2*(x-9)*(x+13)=0

    x=9 , x= -13

    substitute in y=4 +x

                         y= 4 + 9

                         y=13

    13-9=4 checked

    13^2 + 9^2=250

    169 +81=250 checked

    x=9 , y= 13 your two integers

  • a - b = 4

    a^2 + b^2 = 250

    (a - b)^2 = 4^2

    a^2 - 2ab + b^2 = 16

    250 - 2ab = 16

    125 - ab = 8

    117 = ab

    a = 4 + b

    117 = b * (b + 4)

    117 = b^2 + 4b

    117 + 4 = b^2 + 4b + 4

    121 = (b + 2)^2

    -11 , 11 = b + 2

    -13 , 9 = b

    a = 4 + b

    a = -9 , 13

    -13 , -9

    and

    9 , 13

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