Math Question....Factoring?

The sum of the squares of two consecutive positive integers is 145. Find the integers. Please explain step by step...Thanks

6 Answers

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  • 1 decade ago
    Favorite Answer

    x^2 + (x+1)^2 = 145

    x^2 + x^2 + 2x +1 = 145

    2x^2 + 2x -144 = 0

    (2x - 16)(x + 9) = 0

    x = 8 or -9 are your roots, but you want positive integers, so ignore -9.

    x = 8 and x+1 = 9 are the two consecutive positive integers.

  • 1 decade ago

    (8, 9)

    First, call the integers x and x+1. Then, square each and add them together: x^2; x^2 + 2x + 1 yields x^2 + x^2 + 2x + 1 = 2x^2 + 2x + 1. Then set that sum equal to 145 and re-arrange to have the standard quadratic form: 2x^2 + 2x + 1 = 145 (subtract 145 from each side) so: 2x^2 + 2x + 1 - 145 = 145 - 145 so: 2x^2 + 2x - 144 = 0. Then substitute into the quadratic equation's answer form with a = 2, b = 2, and c = -144. This yields two answers, 8 and -9. Since the answer requires two positive consecutive integers, we drop the -9. Since we let x equal the smaller of the two integers, the second one is 9. The answer then is (8, 9). Checking, we find 64 + 81 does equal 145.

  • Puggy
    Lv 7
    1 decade ago

    Let x be one of the two consecutive integers. Being consecutive, it follows that the next one is x + 1.

    The sum of their squares is 145; translating that into an algebraic equation,

    x^2 + (x + 1)^2 = 145

    Solving for x, we get

    x^2 + x^2 + 2x + 1 = 145

    2x^2 + 2x + 1 = 145

    2x^2 + 2x - 144 = 0

    Divide by 2,

    x^2 + x - 72 = 0

    Factor,

    (x + 9)(x - 8) = 0

    Which implies x = {-9, 8}.

    Since we have two solutions for x, it follows that we have two sets of solutions.

    The consecutive numbers are -9 and -8, or 8 and 9.

  • 1 decade ago

    two consecutive integers, x and x+1

    sum of squares (x)^2 + (x+1)^2 equals 145

    x^2 + x^2 + 2x + 1 = 145

    2x^2 + 2x - 144 = 0

    (2x + 18)(x - 8) = 0

    x = 8 or -9 (but you said positive so...)

    the integers are 8 and 9

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  • Amit Y
    Lv 5
    1 decade ago

    Two consecutive positive integers are m and m+1.

    Let's find them knowing that

    m^2 + (m + 1)^2 = 145

    m^2 + m^2 + 2m + 1 = 145

    2m^2 + 2m = 144

    2m^2 + 2m - 144 = 0

    m^2 + m - 72 - 0

    m1,2 = (-1 +- sqrt(1 + 288))/2 = (-1 +- sqrt(289))/2 =

    = (-1 +- 17)/2

    m1 = 16/2 = 8

    m2 = -18/2 = -9 -> Not a solution because it is negative

    8^2 + 9^2 = 64 + 81 = 145

  • John T
    Lv 6
    1 decade ago

    So, you have two positive integers, n and n+1, since they are consecutive.

    n^2 + (n+1)^2 = 145

    n^2 + n^2 + 2n + 1 = 145

    2n^2 + 2n - 144 = 0

    n^2 +n -72 = 0 (multiply through by 1/2 )

    (n+9) (n-8)

    answers are both -9 and 8. There are two solutions.

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