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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
The values of \('a'\) for which one root of the equation \(x^2 - (a +1)\,x + a^2 + a - 8 = 0\) exceeds \(2\) and the other is lesser than \(2\), are given by
- A \(3 < a < 10\)
- B \(a \ge 10\)
- C \(-2 < a < 3\)
- D \(a \le - 2\)
Answer & Solution
Correct Answer
(C) \(-2 < a < 3\)
Step-by-step Solution
Detailed explanation
\(x^{2}-(a+1) x+a^{2}+a-8=0\) Sinceroots are different, therefore \(D>0\) \(\Rightarrow(a+1)^{2}-4\left(a^{2}+a-8\right)>0\) \(\Rightarrow(a-3)(3 a+1)<0\) There are two cases arises. Case \(I\) \(: a-3>0\) and \(3 a+1<0\) \(\Rightarrow a>3\) and \(a<-\frac{11}{3}\) Hence, no…
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