What is |β-α| in terms of sum and product of roots?

|β-α|² = (α+β)² - 4αβ = (sum)² - 4(product). This is Vieta's formula applied to the difference of roots.

How to find integer values of λ?

Square the inequality 1/2 ≤ |β-α| ≤ 3/2 to get 1/4 ≤ 25/9-λ ≤ 9/4. Solve: 19/36 ≤ λ ≤ 91/36. Integer λ: 1 and 2.

Why is the sum 1+2=3?

The only integers in [19/36, 91/36] ≈ [0.527, 2.527] are λ=1 and λ=2. Their sum = 3.

What is the discriminant condition?

For real roots: D≥0 → (20)²-4(12)(3λ)≥0 → 400-144λ≥0 → λ≤25/9≈2.77. Both λ=1,2 satisfy this.

How to square both sides of an inequality?

If 0 ≤ a ≤ b, then a² ≤ b². Since |β-α|≥0 and 1/2,3/2>0, squaring preserves the inequality direction.

What is (β-α)² expanded?

(β-α)² = β²-2αβ+α² = (α+β)²-4αβ = (5/3)²-4(λ/4) = 25/9-λ.

Verify λ=1 satisfies the condition.

|β-α| = √(25/9-1) = √(16/9) = 4/3. Check: 1/2≤4/3≤3/2? 4/3≈1.33 and 3/2=1.5. Yes ✓

Verify λ=2 satisfies the condition.

|β-α| = √(25/9-2) = √(7/9) = √7/3 ≈ 0.882. Check: 1/2≤0.882≤3/2? Yes ✓

Let α,β be roots of 12x²-20x+3λ=0 and 1/2≤|β-α|≤3/2 sum of all λ

Q: What are Vieta's formulas?

For ax²+bx+c=0: sum of roots = -b/a, product of roots = c/a. Here: α+β = 20/12 = 5/3, αβ = 3λ/12 = λ/4.

Q: Why is λ=0 excluded?

λ=0: |β-α|=√(25/9)=5/3≈1.67 > 3/2=1.5. Fails the upper bound condition.

Let α,β be roots of 12x²-20x+3λ=0 and 1/2≤|β-α|≤3/2 sum of all λ | JEE Main Mathematics

NEETJEE

Math Quadratic

Q MCQ Quadratic

Let $\alpha, \beta$ be the roots of the quadratic equation $12x^2 – 20x + 3\lambda = 0,\; \lambda \in \mathbb{Z}$. If $\dfrac{1}{2} \leq |\beta – \alpha| \leq \dfrac{3}{2}$, then the sum of all possible values of $\lambda$ is:

A) 3 ✓

B) 6

C) 1

D) 4

✅ Correct Answer

A) 3

✓

Solution Steps

Apply Vieta’s Formulas

For $12x^2 – 20x + 3\lambda = 0$:

$\alpha + \beta = \dfrac{20}{12} = \dfrac{5}{3}$

$\alpha\beta = \dfrac{3\lambda}{12} = \dfrac{\lambda}{4}$

Express |β − α|² using Vieta

$$|\beta – \alpha|^2 = (\alpha+\beta)^2 – 4\alpha\beta$$

$= \left(\dfrac{5}{3}\right)^2 – 4 \cdot \dfrac{\lambda}{4}$

$= \dfrac{25}{9} – \lambda$

Square the given inequality

$\dfrac{1}{2} \leq |\beta-\alpha| \leq \dfrac{3}{2}$

Squaring (all terms positive):

$$\frac{1}{4} \leq \frac{25}{9} – \lambda \leq \frac{9}{4}$$

Solve for λ

Left inequality: $\dfrac{25}{9} – \lambda \geq \dfrac{1}{4}$

$\lambda \leq \dfrac{25}{9} – \dfrac{1}{4} = \dfrac{100-9}{36} = \dfrac{91}{36} \approx 2.527$

Right inequality: $\dfrac{25}{9} – \lambda \leq \dfrac{9}{4}$

$\lambda \geq \dfrac{25}{9} – \dfrac{9}{4} = \dfrac{100-81}{36} = \dfrac{19}{36} \approx 0.527$

$$\therefore \frac{19}{36} \leq \lambda \leq \frac{91}{36}$$

Find integer values of λ

$\lambda \in \mathbb{Z}$ and $0.527 \leq \lambda \leq 2.527$

Integer values: $\lambda = 1$ and $\lambda = 2$

λ	\|β−α\|²=25/9−λ	\|β−α\|	In [½, 3/2]?
0	25/9 ≈ 2.78	≈ 1.67	❌ > 3/2
1	16/9	4/3 ≈ 1.33	✓
2	7/9	√7/3 ≈ 0.88	✓
3	−2/9 < 0	—	❌ No real roots

Sum all valid λ

Valid values: $\lambda = 1$ and $\lambda = 2$

$$\text{Sum} = 1 + 2 = \boxed{3}$$

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Key Insight

|β−α|² = (sum)²−4(product) = 25/9−λ. Square the bounds to get λ ∈ [19/36, 91/36]. Only integers 1 and 2 qualify.

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Theory

1. Vieta’s Formulas

For $ax^2+bx+c=0$ with roots $\alpha,\beta$: $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$. Here: $\alpha+\beta = 20/12 = 5/3$, $\alpha\beta = 3\lambda/12 = \lambda/4$. These relations connect root properties to coefficients without finding the roots explicitly.

2. Difference of Roots Formula

$(\beta-\alpha)^2 = (\alpha+\beta)^2 – 4\alpha\beta = \dfrac{D}{a^2}$ where $D = b^2-4ac$ is the discriminant. Also $|\beta-\alpha| = \dfrac{\sqrt{D}}{|a|}$. For real and distinct roots we need $D > 0$. Equal roots when $D=0$ give $|\beta-\alpha|=0$.

3. Squaring Inequalities

If $p \leq f(x) \leq q$ with all quantities non-negative, squaring gives $p^2 \leq [f(x)]^2 \leq q^2$. Here $|β-α|\geq 0$ always, and $1/2, 3/2 > 0$, so squaring preserves the inequality direction. This converts the absolute value inequality into a polynomial inequality in $\lambda$.

4. Integer Constraints in JEE Problems

When $\lambda \in \mathbb{Z}$, find the continuous range first, then identify integers in that range. Here $\lambda \in [19/36, 91/36] \approx [0.53, 2.53]$, so only $\lambda = 1$ and $\lambda = 2$. Always verify by substituting back. $\lambda=3$ gives negative discriminant (no real roots), so excluded.

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FAQs

What is |β−α| in terms of discriminant?

⌄

|β−α| = √D / |a| where D = b²−4ac. Here |β−α| = √(400−144λ) / 12 = √(25/9−λ).

What are Vieta’s formulas for this equation?

⌄

α+β = 20/12 = 5/3 and αβ = 3λ/12 = λ/4. From these: |β−α|² = (5/3)² − 4(λ/4) = 25/9 − λ.

How to get λ range from squaring?

⌄

1/4 ≤ 25/9−λ ≤ 9/4. Left: λ ≤ 91/36≈2.527. Right: λ ≥ 19/36≈0.527. So 0.527 ≤ λ ≤ 2.527.

Why is λ=0 excluded?

⌄

λ=0: |β−α|=5/3≈1.67 > 3/2=1.5. Fails upper bound. λ=0 < 19/36 so outside valid range.

Verify λ=1 works.

⌄

|β−α| = √(25/9−1) = √(16/9) = 4/3 ≈ 1.33. Is 1/2 ≤ 1.33 ≤ 1.5? Yes ✓

Verify λ=2 works.

⌄

|β−α| = √(25/9−2) = √(7/9) = √7/3 ≈ 0.882. Is 0.5 ≤ 0.882 ≤ 1.5? Yes ✓

Why is λ=3 excluded?

⌄

λ=3: 25/9−3 = −2/9 < 0. Square root of negative → no real roots. Also λ=3 > 91/36≈2.527.

What is 91/36 in decimal?

⌄

91÷36 = 2.527… So the largest integer ≤ 2.527 is 2. Combined with lower bound 0.527, integers are 1 and 2.

What if λ were not restricted to integers?

⌄

λ ∈ [19/36, 91/36] continuously. Sum would be ∫ but that’s not the question. Integer restriction gives exactly {1,2} with sum 3.

How to compute 25/9 − 9/4?

⌄

LCM of 9,4 is 36. 25/9 = 100/36 and 9/4 = 81/36. So 100/36 − 81/36 = 19/36.

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