Find the value of $2p+4q+7r$ given that $2p,\ q,\ 2r$ are in geometric progression.

Question

It is given that $2p,\ q, \ 2r$ are in G.P.

Also the roots of the quadratic equation $$px^2+qx+r=0$$ are of the form $\alpha ^2,\ 4\alpha -4$.

Find the value of $2p+4q+7r$.

From the given data: Following conclusions can be drawn $$q^2=4pr\ \implies q^2-4pr=0$$

Therefore the given quadratic equation has equal roots. Which gives $$\alpha ^2-4\alpha +4=0$$ i.e. $\alpha=2$

So, we have $$4p+2q+r=0$$.

How do I proceed?

Answer 1

The problem has infinitely many solutions.

As $2p,q$ and $2r$ are in geometric progression, there exists a constant $k$ such that $2pk = q$ and $qk = 2r$.

The discriminant in quadratic equation is $$\Delta = q^2-4pr = q^2-4\frac{q}{2k}\frac{qk}{2}=0.$$ Thus, $x = -\frac{q}{2p} = -k$ is the only solution.

Hence, $\alpha^2 = -k$ and $4\alpha-4=-k \Rightarrow k = -4.$

Thus, $p = -\frac{q}{8}$ and $r = -2q$, vanishing the expression $16p+4q+r$, obtained from the quadratic equation.

Examples:

If $p = -1,q = 8$ and $r = -16$, the sum $2p+4q+7r = -82.$ However, If $p = 2,q = -16$ and $r = 32$, the sum $2p+4q+7r = 164.$

Answer 2

Hint: you have found a quadratic equation with roots $\alpha$, so what could $p,q,r$ be (upto a constant of proportionality)?