Question

The value of $p$ for which $6$ lie between the roots of the equation $x^2+2\left(p-3\right)x+9=0$ is

• A. $\left(\frac{3}{4},\infty \right)$
• B. $\left(-\infty ,\frac{-3}{4}\right)$
• C. $\left(\frac{-3}{4},\frac{3}{4}\right)$
• D. None of these

Answer 1

The correct option is B

$\left(-\infty ,\frac{-3}{4}\right)$

Explanation for the correct answer:

Compute the required value:

Step 1: Write the expression for discriminant of the given equation

Let $f\left(x\right)=x+2\left(p-3\right)x+9=0$

For the roots to be real, the value of the discriminant must be greater than or equal to zero.

For the equation,

$x^2+2\left(p-3\right)x+9=0$, the discriminant is given as

$D=b^2-4ac={\left(2\left(p-3\right)\right)}^2-4\left(9\right)$

Step 2: Apply the condition of discriminants for equation to have real roots

For real roots

$4{\left(p-3\right)}^2-36\ge 0$

$\Rightarrow$ ${\left(p-3\right)}^2-3^2\ge 0$

$\Rightarrow$ $\left(p-3-3\right)\left(p-3+3\right)\ge 0$

$\Rightarrow$ $p\left(p-6\right)\ge 0$

$\Rightarrow$ $p\in \left(-\infty ,0\right)\cup \left(6,\infty \right)$ $...\left(i\right)$

Step 3: Apply the condition for a number to be between the roots

For $6$ to lie between the roots, $f\left(6\right)<0$

$\Rightarrow$ $6^2+12\left(p-3\right)+9<0$

$\Rightarrow$ $45+12p-36<0$

$\Rightarrow$ $12p<-9$

$\Rightarrow$ $p<\frac{-3}{4}$

$\Rightarrow$ $p\in \left(-\infty ,\frac{-3}{4}\right)$ $...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$ we get,

$p\in \left(-\infty ,\frac{-3}{4}\right)$

Hence, option $\left(B\right)$ is the correct answer.