WBJEE · Maths · Quadratic Equation
If \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^{2}+p x+q=0,\) where \(\alpha, \beta, p\) and \(q\) are real, then the roots of the equation \(\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0\) are
- A \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
- B \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)\)
- C \(\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)\)
- D \((\sqrt{\alpha}+\sqrt{\beta})\) and \((\sqrt{\alpha}-\sqrt{\beta})\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)\) and \(\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)\)
Step-by-step Solution
Detailed explanation
Since, \((\alpha+\sqrt{\beta})\) and \((\alpha-\sqrt{\beta})\) are the roots of the equation \(x^{2}+p x+q=0\) \(\therefore \quad\) Sum of roots \(=-p\) \(\Rightarrow \quad(\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-P\) \(\Rightarrow\)…
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