Ratio of Roots of ax²+2bx+c=0 Equals Ratio of Roots of px²+2qx+r=0 — Find b²/q²

A standard quadratic-equations problem, not tied to a specific exam paper.

Problem

If the ratio of the roots of the quadratic equation $ax^2+2bx+c=0$ is the same as the ratio of the roots of $px^2+2qx+r=0$, then $\dfrac{b^2}{q^2}$ is equal to?

Note: the title attached to the worked solution embedded below writes the second equation as $px^2+2px+r=0$, a typo, since $q$ then appears nowhere. The working itself uses $px^2+2qx+r=0$, as here.

Key insight. “Same ratio of roots” for two different quadratics is exactly the kind of condition componendo-dividendo was built for — turning $\dfrac{\alpha_1}{\beta_1}=\dfrac{\alpha_2}{\beta_2}$ into $\dfrac{\alpha_1+\beta_1}{\alpha_1-\beta_1}=\dfrac{\alpha_2+\beta_2}{\alpha_2-\beta_2}$, both sides now built entirely from the sum and product of each equation’s own roots — exactly what Vieta’s formulas hand you directly.

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Approach

Let $\alpha_1,\beta_1$ be the roots of the first equation and $\alpha_2,\beta_2$ the roots of the second, with $\dfrac{\alpha_1}{\beta_1}=\dfrac{\alpha_2}{\beta_2}$. Apply componendo-dividendo to both ratios, then square each side and express everything using $(\alpha+\beta)^2$ and $\alpha\beta$ — both directly available from Vieta’s formulas for each quadratic. Equate the two resulting expressions and simplify.

Solution

Step 1 — Apply componendo-dividendo to the equal ratios

Since $\dfrac{\alpha_1}{\beta_1}=\dfrac{\alpha_2}{\beta_2}$, componendo-dividendo gives:

$$\frac{\alpha_1+\beta_1}{\alpha_1-\beta_1} = \frac{\alpha_2+\beta_2}{\alpha_2-\beta_2}$$

Squaring both sides:

$$\frac{(\alpha_1+\beta_1)^2}{(\alpha_1-\beta_1)^2} = \frac{(\alpha_2+\beta_2)^2}{(\alpha_2-\beta_2)^2}$$

Step 2 — Rewrite (α − β)² using (α + β)² − 4αβ

$$\frac{(\alpha_1+\beta_1)^2}{(\alpha_1+\beta_1)^2-4\alpha_1\beta_1} = \frac{(\alpha_2+\beta_2)^2}{(\alpha_2+\beta_2)^2-4\alpha_2\beta_2}$$

Step 3 — Substitute Vieta’s formulas for each quadratic

For $ax^2+2bx+c=0$: $\alpha_1+\beta_1=-\dfrac{2b}{a}$, $\alpha_1\beta_1=\dfrac{c}{a}$.

For $px^2+2qx+r=0$: $\alpha_2+\beta_2=-\dfrac{2q}{p}$, $\alpha_2\beta_2=\dfrac{r}{p}$.

$$\frac{\dfrac{4b^2}{a^2}}{\dfrac{4b^2}{a^2}-\dfrac{4c}{a}} = \frac{\dfrac{4q^2}{p^2}}{\dfrac{4q^2}{p^2}-\dfrac{4r}{p}}$$

Step 4 — Simplify each side

Multiplying numerator and denominator of the left side by $\dfrac{a^2}{4}$, and the right side by $\dfrac{p^2}{4}$:

$$\frac{b^2}{b^2-ac} = \frac{q^2}{q^2-pr}$$

Step 5 — Cross-multiply

$$b^2(q^2-pr) = q^2(b^2-ac)$$

$$b^2q^2 – b^2pr = b^2q^2 – acq^2$$

Step 6 — Cancel and solve

The $b^2q^2$ terms cancel:

$$-b^2pr = -acq^2 \implies b^2pr = acq^2 \implies \frac{b^2}{q^2} = \frac{ac}{pr}$$

Answer

Common mistakes

• Forgetting the factor of 2 in “2b” and “2q” when applying Vieta’s formulas. For $ax^2+2bx+c=0$, the sum of roots is $-\dfrac{2b}{a}$ (not $-\dfrac{b}{a}$, which would be the formula if the middle term were simply $bx$) — dropping this factor of 2 propagates an error through every subsequent step.
• Not squaring both sides after componendo-dividendo. The raw ratio $\dfrac{\alpha+\beta}{\alpha-\beta}$ still contains a square root implicitly (through $\alpha-\beta$), which doesn’t simplify cleanly using Vieta’s formulas alone — squaring first converts $(\alpha-\beta)^2$ into $(\alpha+\beta)^2-4\alpha\beta$, both directly expressible via sum and product.

Practise next

• If the ratio of the roots of $2x^2+5x+3=0$ equals the ratio of the roots of $3x^2+2qx+8=0$, find $q^2$, using the same componendo-dividendo approach.