Question

The range of $p$ for which $6$ lies between the roots of $x^2+2(p-3)x+9=0$ is

• A. $(-\infty ,0)\cup (6,\infty )$
• B. $(-\infty ,-\frac{3}{4})\cup (6,\infty )$
• C. $(-\infty ,-\frac{3}{4})$
• D. $(-\infty ,-\frac{3}{4}]$

Answer 1

The correct option is C $(-\infty ,-\frac{3}{4})$

Given the quadratic equation $x^2+2(p-3)x+9=0$
Let $\alpha ,\beta$ be the roots of the equation.
Now, $\alpha <6<\beta ,$ which can be shown as:

Now, let $f\left(x\right)=x^2+2(p-3)x+9$
$\Rightarrow f\left(6\right)<0\Rightarrow 36+2(p-3)\times 6+9<0\Rightarrow 45+12p-36<0\Rightarrow 9+12p<0\Rightarrow 12p<-9\Rightarrow p<-\frac{3}{4}\cdots \left(i\right)$

Also, $D>0$
$\Rightarrow 4(p-3{)}^2-36>0\Rightarrow 4(p^2+9-6p)-36>0\Rightarrow p^2-6p+9-9>0\Rightarrow p^2-6p>0\Rightarrow p(p-6)>0\Rightarrow p\in (-\infty ,0)\cup (6,\infty )\cdots (ii)$
From $\left(i\right)\&\left(ii\right),$ we get:
$p\in (-\infty ,-\frac{3}{4})$