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  • The values of   for which one root of the equation <img alt="x^{2}-(a+1)x+a^{2}+a-8=0" src="http://

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 :

Option: 1

3< a< 10


Option: 2

a\geq 10


Option: 3

-2< a< 3


Option: 4

a\leq -2


Answers (1)

best_answer

\begin{aligned} &x^{2}-(a+1) x+a^{2}+a-8=0\\ &\text { Since roots are different, therefore } \mathrm{D}>0\\ &\Rightarrow(a+1)^{2}-4\left(a^{2}+a-8\right)>0\\ &\Rightarrow(a-3)(3 a+11)<0 \end{aligned}

There are two cases that arise

Case 1 a - 3 > 0 and  3a + 11 < 0

\begin{aligned} &\Rightarrow a>3 \text { and } a<-\frac{11}{3}\\ &\text { Hence, no solution in this case } \end{aligned}

Case 2 a - 3 < 0 and  3a + 11 > 0

\\\Rightarrow a<3 \text { and } a>-\frac{11}{3}\\\therefore-\frac{11}{3}<a<3 \Rightarrow-2<a<3

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