Question

Let \(\alpha_1, \alpha_2\) and \(\beta_1, \beta_2\) be the roots of \(ax^2 + bx + c = 0\) and \(px^2 + qx + r = 0\) respectively. If the system of equations \(\alpha_1 y + \alpha_2 z = 0\) and \(\beta_1 y + \beta_2 z = 0\) has a non-trivial solution, then

• A. \(b^2 pr = q^2 ac\)
• B. \(bpr^2 = qac^2\)
• C. \(bpr^2 = qa^2 c\)
• D. None of these

Hint:

Use Vieta's formulas and the condition for non-trivial solutions.

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \begin{vmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{vmatrix} = 0 \Rightarrow \alpha_1\beta_2 = \alpha_2\beta_1 \]
Step 2: Calculation / Simplification}
\(\frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = \frac{\alpha_1+\alpha_2}{\beta_1+\beta_2} = \sqrt{\frac{\alpha_1\alpha_2}{\beta_1\beta_2}}\)
\(\alpha_1+\alpha_2 = -b/a\), \(\alpha_1\alpha_2 = c/a\)
\(\beta_1+\beta_2 = -q/p\), \(\beta_1\beta_2 = r/p\)
\(\frac{-b/a}{-q/p} = \sqrt{\frac{c/a}{r/p}} \Rightarrow \frac{bp}{aq} = \sqrt{\frac{cp}{ar}}\)
Square both sides: \(\frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar} \Rightarrow b^2 p^2 ar = a^2 q^2 cp\)
\(b^2 pr = q^2 ac\)
Step 3: Final Answer
\[ b^2 pr = q^2 ac \]