find three consecutive odd integers such that 3 times the product of the first and third exceeds the product of the first and second by 28. ?

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  • Philip
    Lv 6
    2 weeks ago

    Let 3 consecutive odd integers be 2n-3, 2n-1, 2n+1...(1).

    Then 3(2n-3)(2n+1) = (2n-3)(2n-1) + 28, ie., (2n-3)(6n+3-2n+1) = 28, ie.(2n-3)(n+1) = 7, ie., 2n^2 -n -10 = 0, ie.,   

    (2n-5)(n+2) = 0. Clearly, (1) holds only when n = -2. Then

    the 3 consecutive odd integers are -7, -5, -3.

    Check: 3(2n-3)(2n+1) = 3(-7)(-3) = 63

                 (2n-3)(2n-1) + 28 = (-7)(-5) +28 = 35+28 = 63. 

    Source(s): My source is myself.
  • 3 weeks ago

    a = 2n - 1 ← an odd number cannot be divided by 2

    b = a + 2

    c = b + 2

    3 times the product of the first and third exceeds the product of the first and second by 28.

    3ac = ab + 28

    3.(2n - 1).(b + 2) = (2n - 1).(a + 2) + 28

    3.(2n - 1).(a + 2 + 2) = (2n - 1).(2n - 1 + 2) + 28

    3.(2n - 1).(2n - 1 + 2 + 2) = (2n - 1).(2n - 1 + 2) + 28

    3.(2n - 1).(2n + 3) = (2n - 1).(2n + 1) + 28

    3.(4n² + 6n - 2n - 3) = (4n² + 2n - 2n - 1) + 28

    3.(4n² + 4n - 3) = (4n² - 1) + 28

    12n² + 12n - 9 = 4n² - 1 + 28

    8n² + 12n = 36

    n² + (12/8).n = 36/8

    n² + (3/2).n = 9/2

    n² + (3/2).n + (3/4)² = (9/2) + (3/4)²

    n² + (3/2).n + (3/4)² = (9/2) + (9/16)

    [n + (3/4)]² = 81/16

    n + (3/4) = ± 9/4

    n = - (3/4) ± (9/4)

    n = (- 3 ± 9)/4

    First case: n = (- 3 + 9)/4 = 6/4 = 3/2 ← no possible

    Second case: n = (- 3 - 9)/4 = - 12/4 = - 3 ← ok

    a = - 7

    b = - 5

    c = - 3

  • Como
    Lv 7
    3 weeks ago

    :-

    Let integers be 2x + 1 , 2x + 3 , 2x + 5

    3 ( 2x + 1) ( 2x + 5 ) = ( 2x + 1 ) (2x + 3) + 28

    3 ( 4x² + 12x + 5 ) = 4x² + 8x + 31

    8x² + 28x - 16 = 0

    2x² + 7x - 4 = 0

    ( 2x - 1 ) ( x + 4 ) = 0

    x = - 4 is an integer value

    The three integers are ( -7) , (-5) , (-3)

  • 3 weeks ago

    Three consecutive odd integers are such that

    3 times the product of the first and third

    exceeds the product of the first and second by 28.

    Let (2n – 1), (2n + 1), (2n + 3) be 3 consecutive odd integers. 

    We have 3(2n – 1)(2n + 3) – (2n – 1)(2n + 1) = 28

    12n^2 + 12n – 9 – (4n^2 – 1) – 28 = 0 

    8n^2 + 12n – 36 = 0 

    2n^2 + 3n – 9 = 0 

    (2n – 3)(n + 3) = 0 

    n = –3 or n =  3/2 

    The three consecutive odd integers are -7, -5, -3. 

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  • JOHN
    Lv 7
    3 weeks ago

    Let 2n -1, 2n + 1, 2n + 3 be 3 consecutive odd integers.

    We have 3(2n -1)((2n + 3) – (2n - 1)(2n + 1) – 28 = 0

    12n^2 + 12n – 9 – (4n^2 - 1) – 28 = 0

    8n^2 + 12n – 36 = 0

    2n^2 + 3n – 9 = 0

    (2n - 3)(n + 3) = 0

    n = -3 (the other solution, n =  3/2, is a non-integer

    whinch in any case yields even solutions)

    The numbers are -7, -5, -3.

  • 3 weeks ago

    (2n-1)(2n+3)=3x

    (2n-1)(2n+1)=x+28

     n=-3  x=7

    (-7)(-3)=3(7)

    (-7)(-5)=(7)+28

    LQQD

    (-3),(-5),(-7)

  • 3 weeks ago

    Three consecutive odd integers.  I'll use:

    (x - 2), x, and (x + 2)

    So as long as "x" is odd, the other two will be as well.

    Three times the product of the first and third:

    3(x - 2)(x + 2)

    Exceeds the product of the first and second by 28:

    3(x - 2)(x + 2) = x(x - 2) + 28

    We can simplify both halves and solve as a quadratic:

    3(x² - 4) = x² - 2x + 28

    3x² - 12 = x² - 2x + 28

    2x² + 2x - 40 = 0

    x² + x - 20 = 0

    (x + 5)(x - 4) = 0

    x = 4 and -5

    Since we only want odd values for x we can throw out the 4 to get our answer to be:

    -7, -5, and -3

  • Greg
    Lv 7
    3 weeks ago

    You have a mistake in your question.

  • Gary
    Lv 4
    3 weeks ago

    They are both 8 and 12.

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