Question

If $-(4x+27)<{(x+6)}^2<-4(6+x)$ then find the number of integer value of x

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (A)

0

$-(4x+27)<{(x+6)}^2<-4(x+6)$

${(x+6)}^2>-(4x+27)$ and ${(x+6)}^2<-(4x+27)$

$x^2+12x+36>-4x-27$ and $x^2+12x+36<-4x-24$

$x^2+16x+63>0$ and $x^2+16x+60<0$

$(x+7)(x+9)>0$ and $(x+6)(x+10)<0$

$xϵ(-\infty ,-9)\bigcup (-7,\infty )$ and $xϵ(-10,-6)$

$\therefore xϵ(-10,-9)\bigcap (-7,-6)$

There are zero integer values possible for x.