Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 \mathrm{x}^{2}-31\mathrm{x}+3 \lambda=0$ and $\alpha, \gamma$ be the roots of the equation $35 x^{2}-53 x+4 \lambda=0$. Then $\frac{3 \alpha}{\beta}$ and $\frac{4 \alpha}{\gamma}$ are the roots of the equation :

(A) $7 \mathrm{x}^{2}+245 \mathrm{x}-250=0$

(B) $7 \mathrm{x}^{2}-245 \mathrm{x}+250=0$

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Quadratic Equations and Inequalities

Let

λ \neq = 0

be a real number. Let

α, β

be the roots of the equation

14 x^{2} - 31 x + 3 λ = 0

and

α, γ

be the roots of the equation

35 x^{2} - 53 x + 4 λ = 0

. Then

\frac{3 α}{β}

and

\frac{4 α}{γ}

are the roots of the equation :

Answer: (C)

The shared root

α

satisfies

α + β = \frac{31}{14}, α β = \frac{3 λ}{14}

and

α + γ = \frac{53}{35}, α γ = \frac{4 λ}{35}

. Dividing the products,

\frac{β}{γ} = \frac{15}{8}

, and subtracting the sums,

β - γ = \frac{7}{10}

; together these give

γ = \frac{4}{5}, β = \frac{3}{2}

, and then

α = \frac{5}{7}

. The new roots are

\frac{3 α}{β} = \frac{10}{7}

and

\frac{4 α}{γ} = \frac{25}{7}

, with sum

5

and product

\frac{250}{49}

. The required quadratic is

x^{2} - 5 x + \frac{250}{49} = 0

, that is

49 x^{2} - 245 x + 250 = 0

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option