Question

If $x$ is real, then $3x^2+14x+11>0$, when

• A. $x<-\frac{3}{2}$
• B. $x<\frac{10}{3}$ or $x>1$
• C. $x<-2$
• D. none of the above

Answer 1

The correct option is B $x<\frac{10}{3}$ or $x>1$

${3x}^2+14x+11>0$

$\Rightarrow x^2+\frac{14}{3}x+\frac{11}{3}>0$

$\Rightarrow (x+\frac{10}{3})(x+1)>0$

$\Rightarrow x<-\frac{10}{3}$ or $x>-1$