Quadratic Equation Roots Reciprocal Relation Trick

 

❓ Question

If $\alpha,\beta$ are roots of quadratic equation:

$$\lambda x^2-(\lambda+3)x+3=0$$

and

$$\alpha<\beta$$

such that:

$$\frac1\alpha-\frac1\beta=\frac13$$

then find the sum of all possible values of:

$$\lambda$$

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Properties of roots of quadratic equations
👉 Relation between roots
👉 Sum and product of roots.

Main idea:

$$\frac1\alpha-\frac1\beta = \frac{\beta-\alpha}{\alpha\beta}$$

Use roots formula directly.

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🔷 Step 1 — Use Sum and Product of Roots 💯

Given equation:

$$\lambda x^2-(\lambda+3)x+3=0$$

For quadratic:

$$ax^2+bx+c=0$$

we know:

$$\alpha+\beta=-\frac ba$$
$$\alpha\beta=\frac ca$$

Thus:

$$\alpha+\beta=\frac{\lambda+3}{\lambda}$$
$$\alpha\beta=\frac3\lambda$$

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🔷 Step 2 — Use Given Condition

Given:

$$\frac1\alpha-\frac1\beta=\frac13$$

Using identity:

$$\frac1\alpha-\frac1\beta = \frac{\beta-\alpha}{\alpha\beta}$$

So:

$$\frac{\beta-\alpha}{3/\lambda}=\frac13$$
$$\beta-\alpha=\frac1\lambda$$

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🔷 Step 3 — Use Root Difference Formula

We know:

$$(\beta-\alpha)^2 = (\alpha+\beta)^2-4\alpha\beta$$

Substitute values:

$$\left(\frac1\lambda\right)^2 = \left(\frac{\lambda+3}{\lambda}\right)^2 - 4\left(\frac3\lambda\right)$$

Multiply by:

$$\lambda^2$$
$$1=(\lambda+3)^2-12\lambda$$

Expand:

$$1=\lambda^2+6\lambda+9-12\lambda$$
$$1=\lambda^2-6\lambda+9$$
$$\lambda^2-6\lambda+8=0$$

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🔷 Step 4 — Solve Quadratic

Factorise:

$$(\lambda-2)(\lambda-4)=0$$

Thus:

$$\lambda=2,\ 4$$

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🔷 Step 5 — Find Required Sum

$$2+4=6$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Reciprocal identity galat use kar dena

❌ Root difference formula bhool jaana

❌ $\alpha\beta$ me denominator mistake kar dena

Remember:

$$\boxed{ (\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta }$$

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✅ Final Answer

$$\boxed{6}$$

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📚 Related Topics

• Quadratic Equation Ratio of Roots Problem
• Common Root in Two Quadratic Equations Trick
• Newton’s Theorem Recurrence Relation Shortcut