lf the roots of the equation $\mathrm{a}\mathrm{x}^{2}+\mathrm{b}\mathrm{x}+\mathrm{c}=0$ are the reciprocals of the roots of the equation $\mathrm{p}\mathrm{x}^{2}+\mathrm{q}\mathrm{x}+\mathrm{r}=0$, then
- $\mathrm{a}\mathrm{c}\mathrm{q}^{2}=\mathrm{b}^{2}\mathrm{p}\mathrm{r}$
- $\mathrm{a}\mathrm{c}=\mathrm{p}\mathrm{r}$
- $\mathrm{b}^{2}\mathrm{a}\mathrm{c}=\mathrm{q}^{2}\mathrm{p}\mathrm{r}$
- $\mathrm{a}\mathrm{b}=\mathrm{p}\mathrm{q}$
If roots of ax^2+bx+c=0 are reciprocals of roots of px^2+qx+r=0, then the equation with roots 1/alpha and 1/beta is obtained by swapping coefficients: cx^2+bx+a=0. Comparing this to px^2+qx+r=0, we have c/p = b/q = a/r. From b/q = a/r and b/q = c/p, we derive relations. Specifically, (b/q)^2 = (a/r)*(c/p) = ac/pr, which leads to b^2pr = q^2ac.
If the roots of px^2 + qx + r = 0 are m and n, the roots of ax^2 + bx + c = 0 must be 1/m and 1/n. The sum of the roots for the first equation gives m + n = -q/p and mn = r/p. For the second equation, the sum of the reciprocals is 1/m + 1/n = -b/a and the product of the reciprocals is 1/(mn) = c/a. Since 1/m + 1/n equals (m + n)/(mn), we substitute the values to get (-q/p) / (r/p) = -b/a. Simplifying the left side gives -q/r = -b/a, which rearranges to aq = br. This alone does not perfectly match the given options, but checking the relationship through the product of roots c/a = 1/(r/p) = p/r means ar = cp, which is also not listed. However, we can combine the squared sum ratio: since the sum of roots of the new equation is -b/a, we actually have aq = br, and cross-multiplying the squares of the sum relationship derived from Vieta's formulas yields acq^2 = b^2pr.