preRMO or RMO algebra: my student’s solutions

(These are the solutions from a student of mine, whose identity is private. I will call him, RI, Bengaluru.)

Question I: 

Solve as elegantly as possible: t^{4}=49+20\sqrt{6}

Solution I (of RI, Bengaluru):

t^{4}=49+ 2\times 5 \times \sqrt{24}



t^{2}=5 + 2\sqrt{6}

Hence, t=\sqrt{2}+\sqrt{3}.

Question II: 

Find the necessary and sufficient conditions on the coefficients p, q, and r of the given cubic equation such that the roots of the cubic are in AP:


Solution II: (credit to RI, Bengaluru) : 

Let the roots of the above cubic be a-R, a, a+R.

By Viete’s relations: (a-R)+a +(a+R)=-p and a(a-R)+(a-R)(a+R)+(a+R)(a)=q and a(a-R)(a+R)=r so that we get 3a=p so that a=\frac{p}{3} and from the second equation we get a^{2}-aR+a^{2}-R^{2}+a^{2}+Ra=q, that is, 3a^{2}-R^{2}=q so that a^{2}-R^{2}=q-2\times a^{2} = q - 2 \times (\frac{p}{3})^{2}=q-\frac{2}{9}\times p^{2}, and exploiting the third Viete’s relation we get a \times (a^{2}-R^{2})=r, that is (q - (\frac{2}{9})\times p^{2})a=r, which is the required necessary and sufficient condition, where a=\frac{p}{3}.

Method II: For pedagogical purposes. The above solution to question 2 was quick and elegant because of the right choice of three quantities in AP as a-R, a, a+R. Do you want to know how ugly and messy it can get if a standard assumption is made:

Let the roots of the cubic be \alpha, \beta, \gamma. Let d be the common difference. So that the three roots in AP are \alpha, \beta=\alpha+d and \gamma=\alpha + 2d

Then, applying Viete’s relations, we get \alpha + \beta + \gamma = -p so that 3\alpha + 2d=-p and \alpha\beta+ \beta\gamma + \gamma\alpha=q which changes to \alpha \times (\alpha + d) + (\alpha+d)(\alpha+2d)+ (\alpha+d)(\alpha+2d) = q and the third viete’s relation gives us \alpha (\alpha+d)(\alpha+2d)=r.

The second Viete’s relation is a quadratic in \alpha and the third Viete’s relation is a cubic in \alpha. This is how messy it can get…at least, you will agree RI’s judicious choice has rendered a clean, quick solution.


Nalin Pithwa.



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