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Math Help - The Quadratic Formula

  1. #1
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    Wink The Quadratic Formula

    Using the quadratic formula, find the exact value of x for the following.

    -2kx^2 + 4x + 6k = 0
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by Joker37 View Post
    Using the quadratic formula, find the exact value of x for the following.

    -2kx^2 + 4x + 6k = 0
    The quadratic formula is x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    In your case - assuming k is a constant,

    <br />
a = -2k<br />
b = 4<br />
c = 6k<br />
.

    Just put these into your answer and solve for x. Your answer will have k in it, the "exact value" will most probably involve surds.

    Hope this helps
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Joker37 View Post
    Using the quadratic formula, find the exact value of x for the following.

    -2kx^2 + 4x + 6k = 0
    If you have an equation that says ax^2+bx+c=0 then the quadratic formula is as follows (for k \neq f(x)):

    x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

    • a is whatever is in front of x^2 = -2k
    • b is whatever is in front of x = 4
    • c is whatever does not have an x with it = 6k


    Thus:

    x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}

    Simplify to give x = \frac{1 \mp \sqrt {3k^2+4}}{k}

    There is no way to solve x for a number without knowing the value of k
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  4. #4
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    How did you get from this...

    Quote Originally Posted by e^(i*pi) View Post

    Thus:

    x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}
    ...to this.

    Quote Originally Posted by e^(i*pi) View Post
    Simplify to give x = \frac{1 \mp \sqrt {3k^2+4}}{k}
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Joker37 View Post
    How did you get from this...



    ...to this.

    x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}

    Collecting and multiplying out the terms in the square root gives 16-(-48k^2) = 48k^2+16

    As the HCF of 48 and 16 is 16 we can remove this as a factor: 48k^2+16 = 16(3k^2+1) note, I made a mistake here in my other working - that +3 should have been a +1

    Reintroducing this to the full equation:

    \frac{-4 \pm \sqrt{16(3k^2+1)}}{-4k}

    As \sqrt{ab} = \sqrt{a} \sqrt{b} then \sqrt{16(3k^2+1)} = \sqrt{16} \times \sqrt{3k^2+1} = 4\sqrt{3k^2+1}

    Reintroducing the simplified discriminant:

    \frac{-4 \pm 4\sqrt{3k^2+1}}{-4k} = \frac{4(-1 \pm \sqrt{3k^2+1})}{-4k}

    4 will cancel to give

    \frac{-1 \pm \sqrt{3k^2+1}}{-k}

    You can leave it like that but it's more usual to not have a negative denominator so we multiply by \frac{-1}{-1}

    -\frac{-1 \pm \sqrt{3k^2+1}}{-k} = \frac{1 \mp \sqrt{3k^2+1}}{k}

    Since there is a kČ term under the square root sign then x will be real for all real k
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  6. #6
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    Just to check, \pm and \mp are both the same things, right?

    (This is my first post where I have used LaTex , sort of exciting in a way!).
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  7. #7
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    Quote Originally Posted by Joker37 View Post
    Just to check, \pm and \mp are both the same things, right?

    (This is my first post where I have used LaTex , sort of exciting in a way!).
    The same numbers but in different order!

    Notice that he had first \frac{-1\pm\sqrt{3k^2+1}}{-k} which gives the two solutions \frac{-1+\sqrt{3k^2+1}}{-k}= \frac{1-\sqrt{3k^2+1}}{k} and \frac{-1- \sqrt{3k^2+1}}{-k}= \frac{1+\sqrt{k^2+1}}{k}.

    e^(i*pi) just changed from \pm to \mp to make it clear that the -\sqrt{k^2+1} had changed to +\sqrt{k^2+1} and vice-versa. Of course, it is the same two solutions.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Joker37 View Post
    How did you get from this....
    Joker, check this out. It will help you understand the quadratic formula in it's entirety. Remember, as e-pi has asserted, that generally all constants and variables are considered as coefficient to the quadratic term in question. EG

    2y^3zx^2+yz^4x+19yz=0 is quadratic in x (you just have to see it). To isolate x, use a=2y^3z,b=yz^4,\text{ and }c=19yz.


    http://www.mathhelpforum.com/math-he...c-formula.html
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