Question

The values of k for which each root of the equation , $x^2-6kx+2+2k+9k^2=0$ is greater than 3, always satisfy the inequality :

• A. $7-9k>0$
• B. $11-9k<0$
• C. $29-11k>0$
• D. $29-11k<0$

Answer 1

The correct option is B $11-9k<0$

$f\left(x\right)=x^2-6kx+2-2k+9k^2$

$f\left(x\right)>0$

$\Rightarrow$ $9-18k+2-2k++9k^2>0$

$\Rightarrow$ $9k^2-20k+11>0$

$\Rightarrow$ $9k^2-9k-11k+11>0$

$\Rightarrow$ $9k(k-1)-11(k-1)>0$

$\Rightarrow$ $(k-1)(9k-11)>0$

$\Rightarrow$ $k<1$ or $k>\frac{11}{9}$ ------- ( 1 )

$D\ge 0$

$\Rightarrow$ $36k^2-8+6k-36k^2\ge 0$

$\Rightarrow$ $k-1\ge 0$

$\Rightarrow$ $k\ge 1$ ------- ( 2 )

$\frac{-b}{2a}>3$

$\Rightarrow$ $\frac{6k}{2}>3k>1$ ------ ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

$\Rightarrow$ $9k-11>0$

$\Rightarrow$ $11-9k<0$