Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

One positive integer is 7 less than twice another. The sum of their squares is 145?

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  • 2 months ago

    2x-7=z

    z²+x²=145

    (2x-7)²+x²=145

    (2x-7)(2x-7)+x²=145

    4x²-28x+49+x²=145

    5x²-28x+49=145

    5x²-28x-96=0

    Now use the quadratic equation 

    x=8

    Now plugin x into the first equation to solve for z.

    z=9

  • 2 months ago

    let the intger be  'n' & '2n - 7' 

    Hence 

    n^2 + ( 2n - 7)^2 = 145 

    n^2 + 4n^2 - 28n + 49 = 145 

    5n^2 - 28n - 96 = 0 

    Factor 

    (5n + 12)(n - 8) = 0 

    n- 8 = 0 

    n = 8 

    Hence 

    2n - 7 = 9 

    Hence the integers are 8 & 9 

  • ?
    Lv 7
    2 months ago

    Let x & y be that 2 integers.

    x+7=2y

    x^2+y^2=145

    =>

    y^2+(2y-7)^2=145

    =>

    5y^2-28y-96=0

    =>

    (y-8)(y+12/5)=0

    =>

    y=8 or y=-12/5 (rejected)

    x=2(8)-7=9

    Thus, the integers are

    8 & 9.

  • Ian H
    Lv 7
    2 months ago

    5n^2 - 28n - 96 = (5n + 12)(n - 8)

    Only integer that works is 8

    The sum of their squares is 145 = 64 + 81

    9 is 7 less than twice 8

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  • RR
    Lv 6
    2 months ago

    One positive integer is 7 less than twice another

    x and 2x - 7

    The sum of their squares is 145

    x² + (2x - 7)² = 145

    x² + 4x² - 14x -14 x + 49 = 145

    5x² - 28x + 49 = 145

    5x² - 28x -96 = 0

    (5x + 12) (x - 8) = 0

    5x = - 12 (reject as negative

    x = 8

    if x = 8

    2x - 7 = 16 - 7 = 9

    ANS 8 and 9

  • 2 months ago

    Let the two integers be x and y.

    According to the given condition, 2 y - x = 7

    => x = 2 y - 7 .......... (1)

    Second conditions says : sum of squares of x and y = 145,

    => x² + y² = 145 ..... (2)

    from (1) and (2) we have ---

    => ( 2 y - 7 )² + y² = 145

    => 4 y² - 28 y + y² + 49 = 145

    => 5 y² - 28 y - 96 = 0

    => 5 y² - 40 y + 12 y - 96 = 0

    =>  5y ( y - 8 ) + 12 ( y - 8 ) = 0

    => ( y - 8 ) ( 5 y + 12 ) = 0

    => y = 8 OR y =  - 12/5. Neglect fractional value since y is an iteger.

    => from (1)  x  =  9

    Hence the two Integers are ( 8 and 9 ) ............... Answer

  • ?
    Lv 7
    2 months ago

    Since 13² > 145, we can just work back from 12 and there can't be more than 6 cases to  check.

    145 - 12² = 1, which is square

    (2 x 12) - 1 = 7 ✘ and (2 x 1) - 12 = 7 ✘

    145 - 11² = 24 not square

    145 - 10² = 45 not square

    145 - 9² =  64 = 8²

    8² + 9² = 145 ✓

    (2 x 8) - 9 = 7 ✓

  • 2 months ago

    Excel

    x   2x 2x-7  x^2 (2x-7)^2  Sum

    1    2    -5    1    25            26

    2    4    -3    4    9              13

    3    6    -1    9    1              10

    4    8    1    16    1              17

    5    10    3    25    9             34

    6    12    5    36    25           61

    7    14    7    49    49           98

    8    16    9    64    81         145

    8 & 9

  • 2 months ago

    let x - one positive integer..

     y - other integer

    x = 2y - 7 eq1

    x^2 + y^2 = 145 eq2

    plug in eq1 to eq2 

    (2y - 7)^2 + y^2 = 145

    4y^2 - 28y + 49 +y^2- 145 = 0

    5y^2 - 28y - 96 = 0

     (5y + 12)(y - 8) = 0

    y = - 12/5, y = 8

    solving for x

    when y = 8

    x = 2y - 7

    x = 2(8) - 7 = 9

    when y = - 12/5

    x = 2(-12/5) - 7 = -59/5

    The possible positive integers are 9 and 8...//

  • 2 months ago

    Let x = first positive integer

    Let y = second positive integer

    The first is 7 less than twice the second:

    x = 2y - 7

    The sum of the squares is 145:

    x² + y² = 145

    We now have a system of two equations and two unknowns so we can solve with substitution:

    (2y - 7)² + y² = 145

    4y² - 28y + 49 + y² = 145

    5y² - 28y + 49 = 145

    5y² - 28y - 96 = 0

    Quadratic equation:

    y = [ -b ± √(b² - 4ac)] / (2a)

    y = [ -(-28) ± √((-28)² - 4(5)(-96))] / (2 * 5)

    y = [ 28 ± √(784 + 1920)] / 10

    y = [ 28 ± √(2704)] / 10

    y = (28 ± 52) / 10

    y = -24/10 and 80/10

    y = -12/5 and 8

    Since y is said to be a positive integer we can throw out the negative fraction to leave us with:

    y = 8

    Now we can solve for x:

    x = 2y - 7

    x = 2(8) - 7

    x = 16 - 7

    x = 9

    Your numbers are 9 and 8.

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