Expressions Involving Reciprocal Roots

Given $$p + q = 3$$ and $$p^4 + q^4 = 369$$. Find $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2}$$.

First, we find $$pq$$. We know $$p + q = 3$$, so $$(p+q)^2 = 9 \implies p^2 + q^2 = 9 - 2pq$$. Now, $$(p^2 + q^2)^2 = p^4 + q^4 + 2p^2q^2$$, so $$(9 - 2pq)^2 = 369 + 2(pq)^2$$. Let $$t = pq$$: $$81 - 36t + 4t^2 = 369 + 2t^2$$, which simplifies to $$2t^2 - 36t + 81 - 369 = 0$$ and hence $$2t^2 - 36t - 288 = 0$$. Dividing by 2 gives $$t^2 - 18t - 144 = 0$$, so $$(t - 24)(t + 6) = 0$$. Since $$pq = 24$$ with $$p+q=3$$ leads to complex roots, we take $$pq = -6$$.

Next, we verify that this choice is correct. We have $$p^2 + q^2 = 9 - 2(-6) = 21$$, and therefore $$p^4 + q^4 = 21^2 - 2 \cdot 36 = 441 - 72 = 369$$, as required.

Finally, we compute the desired expression: $$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{3}{-6} = -\frac{1}{2}$$, so $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2} = \left(-\frac{1}{2}\right)^{-2} = 4$$.

Hence the answer is $$4$$.

We are given the quadratic equation in the variable $$x$$

$$3m^{2}x^{2}+m(m-4)x+2=0$$

Let the two (non-zero) roots of this equation be $$\alpha$$ and $$\beta$$. Their ratio is denoted by $$\lambda$$, so we write

$$\lambda=\dfrac{\alpha}{\beta}\qquad\Longrightarrow\qquad \lambda+\dfrac1{\lambda}=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} =\dfrac{\alpha^{2}+\beta^{2}}{\alpha\beta}.$$

The sum and product of the roots of any quadratic $$ax^{2}+bx+c=0$$ are given by the standard relations

$$\alpha+\beta=-\dfrac{b}{a},\qquad \alpha\beta=\dfrac{c}{a}.$$

For our quadratic we have the coefficients

$$a=3m^{2},\qquad b=m(m-4)=m^{2}-4m,\qquad c=2.$$

Therefore

$$\alpha+\beta=-\dfrac{m^{2}-4m}{3m^{2}} =\dfrac{4m-m^{2}}{3m^{2}},$$

$$\alpha\beta=\dfrac{2}{3m^{2}}.$$

To express $$\lambda+\dfrac1{\lambda}$$ in terms of the symmetric functions, first note the algebraic identity

$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta.$$

Using this, we obtain

$$\lambda+\dfrac{1}{\lambda} =\dfrac{\alpha^{2}+\beta^{2}}{\alpha\beta} =\dfrac{(\alpha+\beta)^{2}-2\alpha\beta}{\alpha\beta}.$$

The condition given in the question is

$$\lambda+\dfrac1{\lambda}=1.$$

Substituting the above expression, we get

$$\dfrac{(\alpha+\beta)^{2}-2\alpha\beta}{\alpha\beta}=1.$$

Multiplying both sides by $$\alpha\beta$$ yields

$$(\alpha+\beta)^{2}-2\alpha\beta=\alpha\beta,$$ so that

$$(\alpha+\beta)^{2}=3\alpha\beta.$$

Now we substitute the explicit values of $$\alpha+\beta$$ and $$\alpha\beta$$ obtained from the coefficients.

First compute the square of the sum:

$$(\alpha+\beta)^{2} =\left(\dfrac{4m-m^{2}}{3m^{2}}\right)^{2} =\dfrac{(4m-m^{2})^{2}}{9m^{4}} =\dfrac{m^{2}(4-m)^{2}}{9m^{4}} =\dfrac{(4-m)^{2}}{9m^{2}}.$$

The product is

$$\alpha\beta=\dfrac{2}{3m^{2}}.$$

Setting $$(\alpha+\beta)^{2}=3\alpha\beta$$ gives

$$\dfrac{(4-m)^{2}}{9m^{2}}=3\left(\dfrac{2}{3m^{2}}\right).$$

Simplifying the right-hand side:

$$3\left(\dfrac{2}{3m^{2}}\right)=\dfrac{2}{m^{2}}.$$

Hence we have

$$\dfrac{(4-m)^{2}}{9m^{2}}=\dfrac{2}{m^{2}}.$$

Because $$m\neq0$$, the factors $$m^{2}$$ cancel out, leaving

$$(4-m)^{2}=18.$$

Taking square roots on both sides,

$$4-m=\pm\sqrt{18}=\pm3\sqrt{2}.$$

Thus the possible values of $$m$$ are

$$m=4-3\sqrt{2}\quad\text{or}\quad m=4+3\sqrt{2}.$$

Among these, the least value is clearly

$$m=4-3\sqrt{2}.$$

This value exactly matches option D in the list provided.

Hence, the correct answer is Option D.