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Math Help - find intersepts

  1. #1
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    find intersepts

    I have a math problem that says- Find all of the x-intercepts, if any, of the graph of the function f(x)= 6^2-28x+16

    Could someone tell me step by step how to get the answer to that? Thanks!
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  2. #2
    Senior Member DivideBy0's Avatar
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    (I'm going to assume you meant f(x) = 6x^2 - 28x + 16)

    To find the x-Intercepts, simply find the roots of the equation 6x^2-28x+16=0. There are several ways one can do this. The way I do it is as follows:

    6x^2 - 4x - 24x + 16 = 0 (Splitting the x term up so that the two add to equal -28 and multiply to give 6*16=96)
    2x(3x-2) - 8(3x-2) = 0
    (2x-8)(3x-2) = 0

    Hence the roots are:

    2x-8=0
    2x=8
    x=4

    and

    3x-2=0
    3x=2
    x=2/3

    and therefore the 2 x-Intercepts are (4,0) and (2/3,0).
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  3. #3
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    Wow, thanks very much! And, yea that's what I meant, sorry. I was also wondering what I would do if say the the a and c terms were really large. say, for example -40x^2+49x-15=0

    Thanks!!!
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Wow, thanks very much! And, yea that's what I meant, sorry. I was also wondering what I would do if say the the a and c terms were really large. say, for example -40x^2+49x-15=0

    Thanks!!!
    Using the same method DividedBy0 just used:

    First, I don't like the x^2 term having a negative exponent, so lets divide out a -1 from every term:
    40x^2 - 49x + 15 = 0

    (We need two numbers that add to give -49 and multiply to give 40*15 = 600. Notice that since they multiply to give a positive 600, both signs must be the same, and since 49 is negative, both numbers must be negative: -24 and -25).

    (40x^2 - 24x) + (-25x + 15) = 0
    8x(5x - 3) - 5(5x - 3) = 0
    (8x - 5)(5x - 3) = 0

    So,
    8x - 5 = 0
    8x = 5
    x = 5/8

    And,
    5x - 3 = 0
    5x = 3
    x = 3/5
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  5. #5
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    That was awesome. Is there another way that I can solve this problem. It is taking me forever to find 2 numbers that equal the product of the large AC terms that also equal the B term when added.

    Thanks
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Mathfailure View Post
    That was awesome. Is there another way that I can solve this problem. It is taking me forever to find 2 numbers that equal the product of the large AC terms that also equal the B term when added.

    Thanks
    If you are familiar with the quadratic formula, that is the most obvious way to do it when the numbers become hard to work with (such as if they're too large, they are fractions, or the problem is unfactorable). However, on the question I answered above I used a little trick to find the answer:

    We needed 2 numbers that add to give -49 and multiply to give 600:

    If they multiply to give 600, then we need to find the factors of 600 (but the factors that we need can take a long time to figure out with big numbers like this). Here's the trick:

    I already know that 15 and 40 are factors of 600, right? I can use this fact to find *any* two factors of 600 with a little manipulation.

    15 = 5 x 3
    40 = 8 x 5
    Therefore,
    600 = 15 x 40 = (5 x 3) x (8 x 5)... but multiplication is communative. I can rewrite the order of these numbers:
    600 = (8) x (5 x 5 x 3) = 8 x 75
    600 = (8 x 5 x 3) x (5) = 120 x 5
    600 = (8 x 3) x (5 x 5) = 24 x 25 ... Tada! That's the combination I needed!!!

    Summary: if you have a big number, factor it down to a few of its factors and rearange those factors to find other factors of that number. This should help you find the factors you need.
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  7. #7
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    Thanks very much. I was hoping you could help me with this also.

    The question is Find all real solutions, if any, of the quadratic function.

    2x^2-3x-8=0

    Since -8x2=-16 there is no product that equals -16 and will also add to -3, so what do I do? Thanks!!
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Thanks very much. I was hoping you could help me with this also.

    The question is Find all real solutions, if any, of the quadratic function.

    2x^2-3x-8=0

    Since -8x2=-16 there is no product that equals -16 and will also add to -3, so what do I do? Thanks!!
    For this, you have no choice but to use the quadratic formula. I hope you're familiar with it.
    . - - - . _________
    . . -b b^2 - 4ac
    x = ---------------
    . . . . . . . 2a

    Where
    a = 2
    b = -3
    c = -8

    Plug these values in to solve for the roots of x.
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  9. #9
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Thanks very much. I was hoping you could help me with this also.

    The question is Find all real solutions, if any, of the quadratic function.

    2x^2-3x-8=0

    Since -8x2=-16 there is no product that equals -16 and will also add to -3, so what do I do? Thanks!!
    There is also another method, which is simpler than the quadratic formula when a, b, and c (in ax^2+bx+c=0), are relatively easy and small, as in your problem. It is important to learn how to complete the square because the quadratic formula is derived from it.

    2x^2-3x-8=0

    First, make the coefficient of the x^2 term = 1 (divide by 2)

    x^2-3/2x-4=0

    Now here's the tricky part, recall how if you have (a+b)^2, it expands to a^2+2ab+b^2? Well, in this equation, x^2 is the a^2 term, and 3/2x is the 2ab term. We must 'create' the b^2 term, and to do this, we find half the coefficient of the x term squared:

    x^2 - 3/2x + 9/16 - 9/16 - 4 =0 (We must also subtract it so the equation's value remains the same)

    Now, we can factorise x^2-3/2x+9/16:

    (x - 3/4)^2 - 9/16 - 4 = 0

    (x - 3/4)^2 = 73/16

    And take the square root of both sides:

    x - 3/4 = + sqrt{73}/4
    or
    x - 3/4 = -sqrt{73}/4
    --------------------
    x = (3 - sqrt{73})/4
    or
    x = (3 + sqrt{73}/4

    This is a nice little animation showing how the quadratic formula is derived from completing the square:

    Quadratic formula - derivation
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    There is also another method, which is simpler than the quadratic formula when a, b, and c (in ax^2+bx+c=0), are relatively easy and small, as in your problem. It is important to learn how to complete the square because the quadratic formula is derived from it.

    2x^2-3x-8=0

    First, make the coefficient of the x^2 term = 1 (divide by 2)

    x^2-3/2x-4=0

    Now here's the tricky part, recall how if you have (a+b)^2, it expands to a^2+2ab+b^2? Well, in this equation, x^2 is the a^2 term, and 3/2x is the 2ab term. We must 'create' the b^2 term, and to do this, we find half the coefficient of the x term squared:

    x^2 - 3/2x + 9/16 - 9/16 - 4 =0 (We must also subtract it so the equation's value remains the same)

    Now, we can factorise x^2-3/2x+9/16:

    (x - 3/4)^2 - 9/16 - 4 = 0

    (x - 3/4)^2 = 73/16

    And take the square root of both sides:

    x - 3/4 = + sqrt{73}/4
    or
    x - 3/4 = -sqrt{73}/4
    --------------------
    x = (3 - sqrt{73})/4
    or
    x = (3 + sqrt{73}/4

    This is a nice little animation showing how the quadratic formula is derived from completing the square:

    Quadratic formula - derivation
    well, i don't think completing the square is "easier" than the quadratic formula, but you will have to learn to complete the square eventually, because some problems require it (if we're working with the equation of a circle for instance).

    Nevertheless, it was a good lesson--and the animation was nice too. thanks DivideBy0
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