# A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

$P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

$P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$

$P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$

$P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$

$P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

# A quadratic equation question for pRMO or preRMO

Question:

Find the necessary and sufficient condition that the quadratic equation $ax^{2}+bx+c=0$ where $a \neq 0$ has one root which is the square of the other.

Solution:

Let the two roots of the given quadratic equation $ax^{2}+bx+c=0$, with $a \neq 0$ be $\alpha$ and $\beta$ such that $\beta = \alpha^{2}$.

Then, we know $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$ so that $\alpha+\alpha^{2}=-\frac{b}{a}$ and $\alpha^{3}=\frac{c}{a}$. From the latter relation, we get that $\alpha = (\frac{c}{a})^{\frac{1}{3}}$. Substituting this in the first relation of sum of roots, we get the following necessary and sufficient condition:

$(\frac{c}{a})^{\frac{1}{3}} + (\frac{c}{a})^{\frac{2}{3}} = -\frac{b}{a}$.

The above is the desired solution.

🙂 🙂 🙂

Nalin Pithwa

# A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of

$\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients:

$\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as

$\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get:

$\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that

$\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes:

$(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.

# 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.