Quadratic Equation Question 6
Question 6 - 29 January - Shift 1
Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 x^{2}-31 x+3 \lambda=0$ 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:
(1) $7 x^{2}+245 x-250=0$
(2) $7 x^{2}-245 x+250=0$
(3) $49 x^{2}-245 x+250=0$
(4) $49 x^{2}+245 x+250=0$
Show Answer
Answer: (3)
Solution:
Formula: Sum and Product of Roots
$ 14 \alpha^2-31 \alpha+3 \lambda=0 $
$35 \alpha^2-53 \alpha+4 \lambda=0$ by eliminating $\lambda \Rightarrow \alpha=\frac{5}{7}$
$ \begin{gathered} \alpha+\beta=\frac{31}{14} \\ \beta=\frac{31}{14}-\frac{5}{7}=\frac{21}{14}=\frac{3}{2} \\ \gamma=\frac{53}{35}-\frac{5}{7}=\frac{53-25}{35}=\frac{28}{35}=\frac{4}{5} \\ x^2-\frac{3 \cdot \frac{5}{7}}{\frac{3}{2}}+\frac{4 \cdot \frac{5}{7}}{\frac{4}{5}} x+\frac{12 \cdot \frac{25}{49}}{\frac{3}{2} \cdot \frac{4}{5}}=0 \\ x^2-\frac{10}{7}+\frac{25}{7} x+\frac{250}{49}=0 \\ 49 x^2-245 x+250=0 \end{gathered} $
