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Math Help - problems solving -quadratic equation

  1. #1
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    Lightbulb problems solving -quadratic equation

    Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

    Here goes,

    For which real values of a does the quadratic equation
    F(x)=x^2-ax+12=0
    Have its roots two consecutive integers? What are these roots?

    -im not sure if this will help but I got a as 7??this was throught guess and check
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by phatjigga View Post
    Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

    Here goes,

    For which real values of a does the quadratic equation
    F(x)=x^2-ax+12=0
    Have its roots two consecutive integers? What are these roots?

    -im not sure if this will help but I got a as 7??this was throught guess and check
    Yeah, this is a weird problem. It seems easy at first, but its weird. The following method is not one i'm proud of, but i dont see another way to do this. I tried completing the square, and the quadratic formula to no avail. But here's what i did.

    x^2-ax+12=0

    => (x - m)(x - n) = 0 where m*n = 12 and m + n = a
    the only two consecutive integers that multiply to give 12 is 3 and 4.
    thus a = 3 + 4 = 7

    The reason i dont like this method is because there is guessing involved at the end. with 12 it wasnt so bad, but if our number was a lot larger than 12, we might have had a rough time thinking of two consecutive integers to multiply and get it.

    I'm sure another mathematician will come up with a better way to do this soon
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by phatjigga View Post
    Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

    Here goes,

    For which real values of a does the quadratic equation
    F(x)=x^2-ax+12=0
    Have its roots two consecutive integers? What are these roots?

    -im not sure if this will help but I got a as 7??this was throught guess and check
    Well if the roots are to be integers then the discriminant has to be greater than or equal to 0. So:
    (-a)^2 - 4*1*12 >= 0

    a^2 >= 48

    a >= sqrt(48) = 4*sqrt(3)

    So two real roots are possible.

    What are the roots of the equation in terms of a?
    x = (a (+/-) sqrt(a^2 - 48))/2

    If these two roots are to be consecutive integers then
    x- = (a - sqrt(a^2 - 48))/2
    x+ = (a + sqrt(a^2 - 48))/2
    differ by 1.

    So let (x+) = (x-) + 1

    (a + sqrt(a^2 - 48))/2 = (a - sqrt(a^2 - 48))/2 + 1

    a + sqrt(a^2 - 48) = a - sqrt(a^2 - 48) + 2

    2*sqrt(a^2 - 48) = 2

    sqrt(a^2 - 48) = 1

    a^2 - 48 = 1

    a^2 = 49

    a = (+/-)7

    Now, which one of these work? The easiest way is simply to solve the quadratics:
    a = 7: x^2 - 7x + 12 = 0
    Has solutions x = 3, 4. So a = 7 works.

    a = -7: x^2 + 7x + 12 = 0
    Has solutions x = -3, -4. So a = -7 also works.

    -Dan
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Well if the roots are to be integers then the discriminant has to be greater than or equal to 0. So:
    (-a)^2 - 4*1*12 >= 0

    a^2 >= 48

    a >= sqrt(48) = 4*sqrt(3)

    So two real roots are possible.

    What are the roots of the equation in terms of a?
    x = (a (+/-) sqrt(a^2 - 48))/2

    If these two roots are to be consecutive integers then
    x- = (a - sqrt(a^2 - 48))/2
    x+ = (a + sqrt(a^2 - 48))/2
    differ by 1.

    So let (x+) = (x-) + 1

    (a + sqrt(a^2 - 48))/2 = (a - sqrt(a^2 - 48))/2 + 1

    a + sqrt(a^2 - 48) = a - sqrt(a^2 - 48) + 2

    2*sqrt(a^2 - 48) = 2

    sqrt(a^2 - 48) = 1

    a^2 - 48 = 1

    a^2 = 49

    a = (+/-)7

    Now, which one of these work? The easiest way is simply to solve the quadratics:
    a = 7: x^2 - 7x + 12 = 0
    Has solutions x = 3, 4. So a = 7 works.

    a = -7: x^2 + 7x + 12 = 0
    Has solutions x = -3, -4. So a = -7 also works.

    -Dan
    See i told you, except it wasn't a mathematician, it was a physicist, but close enough. i did what topsquark did up to that "differ by 1" line. for some reason i missed that and so tried another way
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