JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha, \beta\) be the roots of the equation \(x^{2}-\sqrt{2} x+\sqrt{6}=0\) and \(\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1\) be the roots of the equation \(x^{2}+a x+b=0\). Then the roots of the equation \(x ^{2}-( a + b -2) x +( a + b +2)\) \(=0\) are\(...\)
- A non-real complex numbers
- B real and both negative
- C real and both positive
- D real and exactly one of them is positive
Answer & Solution
Correct Answer
(B) real and both negative
Step-by-step Solution
Detailed explanation
\(a=\frac{-1}{\alpha^{2}}-\frac{1}{\beta^{2}}-2\) \(b =\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+1+\frac{1}{\alpha^{2} \beta^{2}}\) \(a + b =\frac{1}{(\alpha \beta)^{2}}-1=\frac{1}{6}-1=-\frac{5}{6}\) \(x^{2}-\left(-\frac{5}{6}-2\right) x+\left(2-\frac{5}{6}\right)=0\)…
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