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  • Let be a real number. Let <img alt="\alpha, \beta" src="https://entrancecorner.oncodecogs.c

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 \mathrm{x}^2-53 \mathrm{x}+4 \lambda=0. Then \frac{3 \alpha}{\beta}\: and\: \frac{4 \alpha}{\gamma} are the roots of the equation
 

Option: 1

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


Option: 2

7 \mathrm{x}^2+245 \mathrm{x}-250=0
 


Option: 3

7 x^2-245 x+250=0

 


Option: 4

\mathrm{49 x^2+245 x+250=0}


Answers (1)

best_answer

14 \mathrm{x}^2-31 \mathrm{x}+3 \lambda=0

\mathrm{and \: 35 x^2-53 x+4 \lambda=0}

Now, one root is common then

\begin{aligned} & \therefore 14 \alpha^2-31 \alpha+3 \lambda=0 \\ \end{aligned}.................(1)

\begin{aligned} & 35 \alpha^2-53 \alpha+4 \lambda=0 \\ \end{aligned}.....................(2)

\begin{aligned} & \frac{\alpha^2}{-124 \lambda+159 \lambda}=\frac{-\alpha}{56 \lambda-105 \lambda}=\frac{1}{343} \\ \end{aligned}

\begin{aligned} & \Rightarrow \frac{\alpha^2}{35 \lambda}=\frac{\alpha}{49 \lambda}=\frac{1}{343} \end{aligned}
\begin{aligned} & \Rightarrow \alpha=\frac{\lambda}{7} \quad\{\text { from (ii) and (iii) } \\ \end{aligned}

\begin{aligned} & \text { and } \alpha^2=\frac{35 \lambda}{343} \\ \end{aligned}

\begin{aligned} & \Rightarrow \frac{\lambda^2}{49}=\frac{35 \lambda}{343} \\ \end{aligned}

\begin{aligned} & \lambda^2-5 \lambda=0 \\ \end{aligned}

\begin{aligned} & \lambda(\lambda-5)=0 \\ \end{aligned}

\begin{aligned} & \lambda=0, \lambda=5 \quad \Rightarrow \alpha=5 / 7 \\ \end{aligned}

\begin{aligned} & \text { not possible } \therefore \text { only } \lambda=5 \text { possible } \\ \end{aligned}

\begin{aligned} & \text { Now, } \alpha+\beta=\frac{31}{14}, \alpha \beta=\frac{3 \lambda}{14}, \alpha+\gamma=\frac{53}{35}, \alpha \gamma=\frac{4 \lambda}{35} \\ \end{aligned}

\begin{aligned} & \therefore \beta=\frac{3}{2} \text { and } \gamma=\frac{4}{5} \end{aligned}
Now equation having roots \left(\frac{3 \alpha}{\beta}, \frac{4 \alpha}{\gamma}\right)=\left(\frac{10}{7}, \frac{25}{7}\right) is

\begin{aligned} & x^2-\frac{35}{7} x+\frac{250}{49}=0 \\ & \Rightarrow 49 x^2-245 x+250=0 \end{aligned}
 

Posted by

Anam Khan

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