Question

Find the minimum value of $\frac{x^2+2x+4}{(x+2)}$ is ?

• A. $2$
• B. $1$
• C. $3$
• D. doesn't exist

Answer 1

The correct option is A $2$

Let, $y=\frac{x^2+2x+4}{x+2}\Rightarrow yx+2y=x^2+2x+4\Rightarrow x^2+2x-yx+4-2y=0\Rightarrow x^2+(2-y)x+(4-2y)=0$

for $x$ to be real, discriminant should be greater than or equal to zero,

$b^2-4ac\ge 0\Rightarrow {(2-y)}^2-4(4-2y)\ge 0\Rightarrow 4+y^2-4y-16+8y\ge 0\Rightarrow y^2+4y-12\ge 0\Rightarrow y^2+6y-2y-12\ge 0\Rightarrow y(y+6)-2(y+6)\ge 0\Rightarrow (y+6)(y-2)\ge 0\Rightarrow y\le -6,y\ge 2$

Hence, the minimum value of $\frac{x^2+2x+4}{x+2}$ is $2.$