Question

The value of a for which one root of the equation $x^2-(a+1)x+a^2+a-8=0$ exceeds 2 and other is less than 2 , are given by

• A. $3<a<10$
• B. $a\ge 10$
• C. $-2<a<3$
• D. $a\le -2$

Answer 1

Answer: (C)

As the roots are real and distinct
$(a+1{)}^2-4\left(a^2+a-8\right)>0$
$3a^2+2a-33<0$
$\Rightarrow (3a+11)(a-3)<0\Rightarrow -\frac{11}{3}<a<3$(1)

Also, for
$\Rightarrow 2^2-2(a+1)+a^2+a-8<0$
$\Rightarrow a^2-a-6<0\Rightarrow (a-3)(a+2)<0$
$\Rightarrow -2<a<3$ (2)
From (1) and (2), we get $-2<a<3$.