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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If \(\alpha \) and \(\beta \) are roots of the equation \(x^2 + px + \frac {3p}{4} = 0,\) such that \(\left| {\alpha - \beta } \right| = \sqrt {10} ,\) then \(p\) belongs to the set
- A \(\{2,\,-5\}\)
- B \(\{-3,\,2\}\)
- C \(\{-2,\,5\}\)
- D \(\{3,\,-5\}\)
Answer & Solution
Correct Answer
(C) \(\{-2,\,5\}\)
Step-by-step Solution
Detailed explanation
Given quadratic eqn. is \(x^{2}+p x+\frac{3 p}{4}=0\) So, \(\alpha+\beta=-p, \alpha \beta=\frac{3 p}{4}\) Now, given \(|\alpha-\beta|=\sqrt{10}\) \(\Rightarrow \alpha-\beta=\pm \sqrt{10}\) \(\Rightarrow(\alpha-\beta)^{2}=10\)…
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