JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha \) and \(\beta \) be the roots of equation \(p{x^2} + qx + r = 0\) ( where \(p \ne 0\)) . If \(p,q,r\) are in \(A.P.\) and \(\frac{1}{\alpha } + \frac{1}{\beta } = 4\) , then the value of \(\left| {\alpha - \beta } \right| \) is
- A \(\frac{{\sqrt {61} }}{9}\)
- B \(\frac{{2\sqrt {17} }}{9}\)
- C \(\frac{{\sqrt {34} }}{9}\)
- D \(\frac{{2\sqrt {13} }}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{{2\sqrt {13} }}{9}\)
Step-by-step Solution
Detailed explanation
since, \(\alpha \) and \(\beta \) be the roots of equation \(p x^{2}+q x+r=0, p \neq 0\) \(\therefore \quad \alpha+\beta=\frac{-q}{p}, \alpha \beta=\frac{r}{p}\) since, \(p, q\) and \(r\) are in \(A P\) \(\therefore \quad 2 q=p+r\) Also, \(\frac{1}{\alpha}+\frac{1}{\beta}=4\)…
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