Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

If 113 is a prime number, how would you factor x^2+2x-113=0?

One of my algebra assignments was to find two consecutive integers such that the sum of their square is 113. I made the equation x(x+2)=113 and distributed, and got x^2+2x=113. Then I subtracted 113 from each side to get the Zero-Product property. But now, the only factors of 113 is 113 x 1, and that doesn't add up to be 2. Please help! What am I doing wrong?

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  • 1 decade ago
    Favorite Answer

    Well, let's start over. To find two consecutive integers such that the sum of their squares is 113, we use the equation

    n² + ( n + 1 )² = 113

    Simplifying and solving, we see that

    n² + n² + 2n + 1 = 113

    2n² + 2n + 1 = 113

    2n² + 2n - 112 = 0

    n² + n - 56 = 0

    ( n - 7) ( n + 8 ) = 0

    The positive root is 7, so the answer must be 7 and 8. Sure enough,

    7² + 8² = 49 + 64 = 113

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  • 1 decade ago

    Your setup is incorrect. The sum of squares means you must square first, then sum. If n is the first integer, n+1 is the next, or 1st consecutive, integer. The problem then asks, for what n does the following hold:

    n^2 + (n+1)^2 = 113

    The solution follows:

    2n^2 + 2n + 2 = 113

    n^2 + n - 56 = 0

    (n+8)(n-7) = 0

    Thus, the answer is 7 and -8 since 7^2 + 8^2 = 49 + 64 = 113 and (-8)^2 + (-7)^2 = 113

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  • Pamela
    Lv 6
    1 decade ago

    Let's build an equation that goes with the words. For consecutive integers x and x +1, their squares are x² and (x + 1)²

    The sum of their squares is x² + (x + 1)² = 113. Simplify.

    x² + x² + 2x + 1 = 113

    2x² + 2x - 112 = 0

    Divide by 2

    x² + x - 56 = 0

    Factor using FOIL

    (x + 8)(x - 7) = 0

    x = -8 and x = 7

    For x = - 8, the next integer is -7.

    (-8)² + (-7)² = 64 + 49 = 113

    For x = 7, the next integer is 8

    (7)² + (8)² = 49 + 64 = 113

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  • Anonymous
    1 decade ago

    If the two consecutive integers are x and x+1 then the sum of the squares is x² + (x+1)² so the equation is:

    x² + (x+1)² = 113

    x² + x² +2x + 1 = 113

    2x² + 2x + 1 = 113

    2x² + 2x - 112 = 0

    Dividing by 2:

    x² + x - 56 = 0

    (x+8)(x-7) = 0

    x+8=0 or x-7=0

    x=-8 or x=7

    So there are two possible solutions:

    the two integers are -8 and -7

    or they are 7 and 8.

    I hope this helps.

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  • Anonymous
    4 years ago

    First step is get all terms on the left by subtracting 24 from each side -- setting the equation equal to zero. x² - 2x - 24 = 0 Some equations can be factored easily, for others use the quadratic equation. Think of 2 numbers whose product is 24 and differ by 2 (from the 2x) Don't worry about signs yet. The numbers are 6 and 4. [If you can't think of any, time for the quadratic equation.] (x - 6)(x + 4) = 0 Since -24 is negative, you know one factor is + and one -. Since -2x is negative, you know the larger factor will be -.

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  • 1 decade ago

    two consecutive integers x , x + 1

    the sum of their square is 113..........x^2 + (x + 1 )^2 = 113

    x^2 + x^2 + 2x + 1 = 113

    2x^2 + 2x - 112 = 0

    x^2 + x - 56 = 0

    (x+8)(x-7)=0

    x = -8 or x = 7

    the 2 numbers are -8 and -7 OR 7 and 8

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  • 1 decade ago

    x(x + 1)=113, first mistake, two integers are x, x+1, not x +2

    second mistake, squared each number, and added them up:

    x^2 + (x + 1)^2 =113,

    x^2 + x^2 + 2x + 1=113

    x^2 + x - 56 = 0

    (x + 8) (x -7)=0

    x= -8

    x=7

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  • 1 decade ago

    x^2 + 2x - 113 = 0

    x1 = -11.677

    x2 = 9,677

    (x - 9.677)(x - 11.677)

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