Question:
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 :
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 :
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If 2 lies strictly between the roots of a quadratic with positive leading coefficient, then f(2) must be negative.
Updated On: Oct 31, 2026
- \(3\lt a\lt 10\)
- \(a\ge 10\)
- \(-2\lt a\lt 3\)
- \(a\le -2\)
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The Correct Option is C
Solution and Explanation
Let \(f(x)=x^2-(a+1)x+a^2+a-8\). The leading coefficient is positive, so 2 lies between the roots iff \(f(2)\lt 0\).
\(f(2)=4-2(a+1)+a^2+a-8=a^2-a-6\).
\(a^2-a-6\lt 0\Rightarrow (a-3)(a+2)\lt 0\Rightarrow -2\lt a\lt 3\).
(The discriminant is then automatically positive, since f takes a negative value.)
Answer: (3) \(-2\lt a\lt 3\)
\(f(2)=4-2(a+1)+a^2+a-8=a^2-a-6\).
\(a^2-a-6\lt 0\Rightarrow (a-3)(a+2)\lt 0\Rightarrow -2\lt a\lt 3\).
(The discriminant is then automatically positive, since f takes a negative value.)
Answer: (3) \(-2\lt a\lt 3\)
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