# 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):

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

$t^{4}=(5+\sqrt{24})^{2}$

$t^{2}=5+\sqrt{24}$

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

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:

$P(t)=t^{3}+pt^{2}+qt+r$

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.

Regards,

Nalin Pithwa.

This site uses Akismet to reduce spam. Learn how your comment data is processed.