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Math Help - How to solve a quadratic in a non quadratic form

  1. #1
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    How to solve a quadratic in a non quadratic form

    How to Solve a quadratic equation in a non quadratic form such as :

    s + 2abs + b s = 4
    a = 5
    b = 2

    Do I solve the solution by completing the square?

    My tutor has advised me to substitute some values into the equation which I have done shown above, with these values it = 100

    He said also to re arrange and maybe move the b to the right hand side s + 2abs = -b

    Do you know why he has advised me to do this?

    As you can see im kinda hitting a brick wall with this quadratic so any help would be much appreciated :O)

    Richard
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  2. #2
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    Quote Originally Posted by ric115 View Post
    How to Solve a quadratic equation in a non quadratic form such as :

    s + 2abs + b s = 4
    a = 5
    b = 2

    Do I solve the solution by completing the square?

    My tutor has advised me to substitute some values into the equation which I have done shown above, with these values it = 100

    He said also to re arrange and maybe move the b to the right hand side s + 2abs = -b

    Do you know why he has advised me to do this?

    As you can see im kinda hitting a brick wall with this quadratic so any help would be much appreciated :O)

    Richard
    What is your actual equation? Remember there needs to be an equals sign in it

    I would get everything on one side and then use the quadratic formula:
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  3. #3
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    This is the original equation sorry my message wasnt very clear.

    s + 2abs + b

    s = 4
    a = 5
    b = 2

    My tutor has told me to move the b to the right hand side

    s + 2abs = -b

    Could i substitue these values into the equation and have

    16 + 80 + 4 = 100


    Thanks

    Richard
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  4. #4
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    Quote Originally Posted by ric115 View Post
    This is the original equation sorry my message wasnt very clear.

    s + 2abs + b

    s = 4
    a = 5
    b = 2

    My tutor has told me to move the b to the right hand side

    s + 2abs = -(b)

    Could i substitue these values into the equation and have

    16 + 80 + 4 = 100


    Thanks

    Richard
    Given what you said I'll assume it is equal to 0 then. If you move b over to the right hand side as said you can complete the square:

    s^2 + 2abs = -b^2

    Looking at the LHS: s^2+2abs = (s+ab)^2 - {\color{blue}a^2b^2}

    Move the ab over to the other side:

    (s+ab)^2 = a^2b^2-b^2 = b^2(a^2-1)

    Can you go on from there?

    Personally I still think using the quadratic equation would be easier with a=1, b= 2ab, c=b^2
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  5. #5
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    This may sound silly but if i was to use the quadratic equation what would s = ?
    I didnt realise i could use the quadratic formula as i the equation had two square values

    Cheers
    Richard
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by ric115 View Post
    This may sound silly but if i was to use the quadratic equation what would s = ?
    I didnt realise i could use the quadratic formula as i the equation had two square values

    Cheers
    Richard
    You can use it because b^2 is a constant in terms of s (ie: there is no s term there)

    Let a =p and b=q

    s + 2abs + b = s + 2pqs + q^2 = 0 (this will make the next step easier to understand),

    In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively)

    <br />
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2}

    Looking at \sqrt{4p^2q^2 - 4q^2} we can see that 4q is a factor so we can remove that to give \sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1}

    <br />
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}  = \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1})

    If we reintroduce a and b for p and q we get a solution of

    s = b(-2{\color{red}a}\pm \sqrt{a^2-1})

    From there you put in your values of a and b
    Last edited by e^(i*pi); May 5th 2009 at 01:09 PM.
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  7. #7
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    Thank you very much that makes much more sense

    Richard
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  8. #8
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    Quote Originally Posted by ric115 View Post
    Thank you very much that makes much more sense

    Richard
    Just to let you know I made a mistake in my previous work but is now corrected in red
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  9. #9
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    yeah got it

    cheers again
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  10. #10
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    Quote Originally Posted by ric115 View Post
    yeah got it

    cheers again
    In that case it was a test to see if you spotted it
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  11. #11
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    Quote Originally Posted by e^(i*pi) View Post
    You can use it because b^2 is a constant in terms of s (ie: there is no s term there)

    Let a =p and b=q

    s + 2abs + b = s + 2pqs + q^2 = 0 (this will make the next step easier to understand),

    In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively)

    <br />
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2}

    Looking at \sqrt{4p^2q^2 - 4q^2} we can see that 4q is a factor so we can remove that to give \sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1}

    <br />
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}  = \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1})

    If we reintroduce a and b for p and q we get a solution of

    s = b(-2{\color{red}a}\pm \sqrt{a^2-1})

    From there you put in your values of a and b

    hi

    Regarding the formula shown below i have gone through this but theres only one part i dont understand. the last bit of the equation. why do you put brackets around s = 2q(-2p √b - 1) or should the formula be
    s = 2q(-2pq √b - 1)





    I hope you can follow what i ve written lol

    Cheers again

    Richard
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