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Step II 1999, q2

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    I've done the question, but I have no idea how many marks I would get because of formality/rigour/correct use of signs etc.

    If someone could give me some general guidance & marks out of 20 I would appreciate it hugely. It's just so I don't walk naively into the exam in about 2 months and lose huge marks on silly things, cheers.






    Here is a given solution:

    nx^2+2x\sqrt{pn^2+q}+rn+s=0 p\not=r, p>0\ and\ n\in \Bbb{Z}^{+}

    i) Where p=3 q=50 r=2 and s=15 find the set of values for n where the equation above has no real roots.

    For non-real roots we need 4(pn^2+q)-4n(rn+s)<0 \Leftrightarrow pn^2+q<n^2r+ns \Leftrightarrow n^2(p-r)-ns+q<0
    Inserting the given values of p, q, r and s:
    n^2-15n+50<0 \Leftrightarrow (n-\frac{15}{2})^2-\frac{25}{4}<0\Rightarrow \pm (n-\frac{15}{2})<\frac{5}{2}
    Solving for the positive case we have n<10 and the negative case -n<\frac{1}{2}(5-15) \Leftrightarrow n>5
    I.e. the equation lacks real roots when n>5 and n<10

    ii) Prove that if p<r and 4q(p-r)&gt;s^2 then the above equation has no real roots fo any n.

    Recall from i) that for non-real roots n^2(p-r)-ns+q&lt;0
    Solving for n gives n= \frac{s^2\pm\sqrt{s^2-4q(p-r)}}{2(p-r)} and \sqrt{s^2-4q(p-r)} is obviously complex when 4q(p-r)&gt;s^2 and thus there are no real solutions for any n.


    iii) Find when the above equation has real roots when n=1, (p-r)=1 and q=\frac{s^2}{8}
    For real roots (p-r)n^2-ns+q\geq0, and with the above values this transfers to finding where 1-s+\frac{s^2}{8}\geq0 \Leftrightarrow (\frac{s}{2\sqrt{2}}-\sqrt{2})^2\geq1 \Rightarrow \pm(\frac{s}{2\sqrt{2}}-\sqrt{2})\geq1
    Solving the positive case gives s\geq2\sqrt{2}(1+\sqrt{2})=4+2\s  qrt{2} as desired and the negative case gives -s\geq2\sqrt{2}(1-\sqrt{2})=2\sqrt{2}-4\Leftrightarrow s\geq4-2\sqrt{2} as desired.

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Updated: May 9, 2012
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