Common Root in Two Quadratic Equations Trick

 

❓ Question

Let:

$$\lambda \ne 0$$

If $\alpha,\beta$ are roots of:

$$x^2-x+2\lambda=0$$

and $\alpha,\gamma$ are roots of:

$$3x^2-10x+27\lambda=0$$

then find:

$$\frac{\beta\gamma}{\lambda}$$

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🖼 Question Image

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✍️ Short Concept

This question is based on:

👉 Properties of roots
👉 Common root concept
👉 Elimination method.

Main idea:

$$\alpha$$

is common root in both equations.

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🔷 Step 1 — Relations from First Equation 💯

For:

$$x^2-x+2\lambda=0$$

Roots are:

$$\alpha,\beta$$

So:

$$\alpha+\beta=1$$
$$\alpha\beta=2\lambda$$

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🔷 Step 2 — Relations from Second Equation

For:

$$3x^2-10x+27\lambda=0$$

Roots are:

$$\alpha,\gamma$$

Thus:

$$\alpha+\gamma=\frac{10}{3}$$
$$\alpha\gamma=\frac{27\lambda}{3}=9\lambda$$

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🔷 Step 3 — Use Common Root Trick

Since $\alpha$ satisfies both equations:

From first:

$$\alpha^2-\alpha+2\lambda=0$$

Multiply by 3:

$$3\alpha^2-3\alpha+6\lambda=0$$

Second equation:

$$3\alpha^2-10\alpha+27\lambda=0$$

Subtracting:

$$7\alpha-21\lambda=0$$
$$\alpha=3\lambda$$

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🔷 Step 4 — Find $\beta$ and $\lambda$

Using:

$$\alpha\beta=2\lambda$$
$$(3\lambda)\beta=2\lambda$$
$$\beta=\frac{2}{3}$$

Now:

$$\alpha+\beta=1$$
$$3\lambda+\frac23=1$$
$$3\lambda=\frac13$$
$$\lambda=\frac19$$

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🔷 Step 5 — Find $\gamma$

Using:

$$\alpha\gamma=9\lambda$$
$$(3\lambda)\gamma=9\lambda$$
$$\gamma=3$$

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🔷 Step 6 — Final Calculation

$$\frac{\beta\gamma}{\lambda} = \frac{\left(\frac23\right)(3)}{\frac19}$$
$$= \frac{2}{1/9}$$
$$=18$$

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✅ Final Answer

$$\boxed{18}$$

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⭐ Golden JEE Insight

Whenever one root is common in two quadratics:

👉 Write both equations in terms of that root

👉 Eliminate $x^2$ term directly

This gives the common root very fast.