Question

If $\alpha ,\beta$ are the roots of the quadratic equation $x^2+2(k-3)x+9=0$ $(\alpha \ne \beta )$. If $\alpha ,\beta \in (-6,1)$ then find the value of $k.$

• A. $2,9$
• B. $(6,\frac{27}{4})$
• C. $6,\infty$
• D. $(-2,\frac{27}{4})$

Answer 1

The correct option is B $(6,\frac{27}{4})$

Let $f\left(x\right)=x^2+2(k-3)x+9$
$\alpha ,\beta \in (6,-1)$


Conditions :
(i) $D>0$

$\Rightarrow \left\{2\right(k-3){\}}^2-\mathrm{4.1.9}>0$

$\Rightarrow 4(k-3{)}^2-36>0$

$\Rightarrow 4k^2-24k+36-36>0$

$\Rightarrow k(k-6)>0$

$k\in (-\infty ,0)\cup (6,\infty )$

(ii) $f(-6)>0$

$\Rightarrow (-6{)}^2+2(k-3).(-6)+9>0$

$\Rightarrow 36-12k+36+9>0$

$\Rightarrow -12k+81>0\Rightarrow 12k<81$

$\Rightarrow k<\frac{81}{12}\Rightarrow k<\frac{27}{4}$

(iii) $f\left(1\right)>0$

$\Rightarrow 1+2k-6+9>0$

$\Rightarrow 2k+4>0$

$\Rightarrow k>-2$

(iv) $-6<-\frac{b}{2a}<1$

$\Rightarrow -6<(3-k)<1$

$2<k<9\Rightarrow k\in (2,9)$

Taking intersection of above condition, we get

$k\in (6,\frac{27}{4})$