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    The quadratic x2+4x-3=0 was misprinted as x2+4x-3=0 on a worksheet of mine. I have now been set it for homework after expressing interest in it.
    Can anyone offer me advice on solving it algebraically, graphically by plotting y=4x and y=3-x2 I can see there are two solutions.
    Thanks in advance!
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    (Original post by ThreePoint14159)
    The quadratic x2+4x-3=0 was misprinted as x2+4x-3=0 on a worksheet of mine. I have now been set it for homework after expressing interest in it.
    Can anyone offer me advice on solving it algebraically, graphically by plotting y=4x and y=3-x2 I can see there are two solutions.
    Thanks in advance!
    You are going to need to use the graphical method you used or a numerical method.
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    The iterative formula:

    \displaystyle x_{n+1}=\frac{\ln(3-x_n^2)}{\ln 4} converges to the positive root fairly quickly.

    The iterative formula:

    \displaystyle x_{n+1}=\frac{3-4^{x_n}}{x_n} converges to the negative root incredibly slowly.

    Newton Raphson finds either root so long as you start near enough:

    \displaystyle x_{n+1}=x_n - \frac{{x_n}^2 + 4^{x_n}-3}{2x_n + 4^{x_n} \ln 4}
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    (Original post by Mr M)
    The iterative formula:

    \displaystyle x_{n+1}=\frac{\ln(3-x_n^2)}{\ln 4} converges to the positive root fairly quickly.

    The iterative formula:

    \displaystyle x_{n+1}=\frac{3-4^{x_n}}{x_n} converges to the negative root incredibly slowly.
    Great! Thanks very much.
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    (Original post by ThreePoint14159)
    Great! Thanks very much.
    You are welcome. Just checking you know how to use an iterative formula?
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    (Original post by Mr M)
    You are welcome. Just checking you know how to use an iterative formula?
    I do, incidentally just been doing Newton-Raphson in FP1
    I had considered doing numerical methods, just took me a while to differentiate 4x for Newton Raphson but I'm sorted now!
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    (Original post by ThreePoint14159)
    I do, incidentally just been doing Newton-Raphson in FP1
    I had considered doing numerical methods, just took me a while to differentiate 4x for Newton Raphson but I'm sorted now!
    Good.
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    Here's a nice way to prove there are no integral solutions:

    (*) 3-x^2=4^x.

    Case 1: If x<-1 then LHS<0 but RHS is always positive, hence no solutions.

    Case 2: If x=-1 then 2=1/2, contradiction.

    Case 3: If x=0 then 3=1, contradiction.

    Case 4: If x is at least 1 then the RHS has at least one factor of 4. Therefore taking mod 4 of both sides yields 3-x^2\equiv 0\iff x^2\equiv 3\pmod 4, which is absurd since 3 isn't a quadratic residue modulo 4.
 
 
 
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