Question

The range of values of $n$ for which one root of the equation $x^2-(n+1)x+n^2+n-8=0$ is greater than $2$ and the other less than $2$

• A. $(-2,3)$
• B. $(2,3)$
• C. $(-3,2)$
• D. $(-3,3)$

Answer 1

The correct option is A $(-2,3)$

Given: $x^2-(n+1)x+n^2+n-8=0$
Let $\alpha ,\beta$ be the roots of given equation
So, $\alpha <2<\beta$
Now, draw the graph for $f\left(x\right)=x^2-(n+1)x+n^2+n-8$,

Applicable Conditions:
i) $D>0$
ii) $f\left(k\right)<0$
Now,
Condition: (i) $D>0\Rightarrow b^2-4ac>0$
$\Rightarrow (n+1{)}^2-4(1\left)\right(n^2+n-8)>0$
$\Rightarrow n^2+2n+1-4n^2-4n+32>0$
$\Rightarrow -3n^2-2n+33>0$
$\Rightarrow 3n^2+2n-33<0$
$\Rightarrow (3n+11)(n-3)<0$
$\Rightarrow n\in (-\frac{11}{3},3)\cdots \left(1\right)$

Condition: (ii) $f\left(k\right)<0\Rightarrow f\left(2\right)<0$
$f\left(2\right)=2^2+(n+1)2+n^2+n-8<0$
$\Rightarrow n^2-n-6<0$
$\Rightarrow (n+2)(n-3)<0$
$\Rightarrow n\in (-2,3)\cdots \left(2\right)$
$\therefore$ Range of $n:$
$\left(1\right)\cap \left(2\right)=(-\frac{11}{3},3)\cap (-2,3)$
$\Rightarrow n\in (-2,3)$