Let 0 be a real number. Let be the roots of the equation 14 x 2 - 31 x + 3 = 0 and be the roots of the equation 35 x 2 - 53 x + 4 = 0 . Then 3 and 4 are the roots of the equation [2023]

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Solution

Q.
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 [2023]

1 $7 x^{2} + 245 x - 250 = 0$

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

3 $49 x^{2} - 245 x + 250 = 0$

4 $7 x^{2} - 245 x + 250 = 0$

Ans.
(3)

$Root α will satisfy the equations,$

$14 α^{2} - 31 α + 3 λ = 0$ $...(i)$

and $35 α^{2} - 53 α + 4 λ = 0$ $...(ii)$

$Now, equation (i) \times 5 - (ii) \times 2 \Rightarrow 49 α - 7 λ = 0 \Rightarrow α = \frac{λ}{7}$

$Put α = \frac{λ}{7} in equation (i): 14 {(\frac{λ}{7})}^{2} - 31 (\frac{λ}{7}) + 3 λ = 0$

$\Rightarrow λ = 0, 5$

$Since λ \neq 0, so λ = 5$

$∴$ $α = \frac{5}{7} \Rightarrow 7 α - 5 = 0$

$So, other root β = \frac{3}{2} \Rightarrow γ = \frac{4}{5}$

$Then, \frac{3 α}{β} = \frac{3 \times \frac{5}{7}}{\frac{3}{2}} = \frac{30}{21} = \frac{10}{7}, and \frac{4 α}{γ} = \frac{4 \times \frac{5}{7}}{\frac{4}{5}} = \frac{25}{7}$

$So, equation can be written as:$

$x^{2} - (\frac{10}{7} + \frac{25}{7}) x + \frac{10}{7} \times \frac{25}{7} = 0$

$\Rightarrow x^{2} - 5 x + \frac{250}{49} = 0 \Rightarrow 49 x^{2} - 245 x + 250 = 0$

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