If the ratio of the roots of $x^{2} + bx + c = 0$ and $x^{2} + qx + r = 0$ are the same, then

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$r^{2} c = {qb}^{2}$

$r^{2} b = {qc}^{2}$

${rb}^{2} = {cq}^{2}$

${rc}^{2} = {bq}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} α, β a r e r o o t s o f x^{2} + b x + c = 0 t h e n \\ α + β = - b, α β = c \\ L e t γ, δ a r e r o o t s o f x^{2} + q x + r = 0 \\ γ + δ = - q, γ δ = r \\ G i v e n \frac{α}{β} = \frac{γ}{δ} \\ \Rightarrow \frac{α + β}{α - β} = \frac{γ + δ}{γ - δ} \\ \Rightarrow \frac{{(α + β)}^{2}}{{(α - β)}^{2}} = \frac{{(γ + δ)}^{2}}{{(γ - δ)}^{2}} \\ \Rightarrow \frac{{(α + β)}^{2}}{{(α + β)}^{2} - 4 α β} = \frac{{(γ + δ)}^{2}}{{(γ + δ)}^{2} - 4 γ δ} \\ \Rightarrow \frac{b^{2}}{b^{2} - 4 c} = \frac{q^{2}}{q^{2} - 4 r} \\ \Rightarrow b^{2} q^{2} - 4 b^{2} r = q^{2} b^{2} - 4 q^{2} c \\ \Rightarrow b^{2} r = q^{2} c \end{array}$