Question

If the minimum value of $\frac{(a+x)(b+x)}{(c+x)}(x>-c)$, for $a>c$, $b>c$ is $(\sqrt{(a-c)}+\sqrt{(b-c)}{)}^2$, then

• A. $x$ is imaginary
• B. $x$ is real
• C. $x$ is zero
• D. Cannot say

Answer 1

The correct option is B $x$ is real

let y = $\frac{(a+x)(b+x)}{(c+x)},x>-c,a>c,b>c,yϵRcy+xy=x^2+(a+b)x+ab{x}^2+(a+b-y)x+(ab-cy)=0$

let us assume that $x$ is real,

$\therefore {(a+b-y)}^2-4(ab-cy)\ge 0y^2-2y(a+b)+(a+b{)}^2-4ab+4cy\ge 0y^2-2y(a+b-2c)+(a-b{)}^2\ge 0$

roots of above equation are $y=\frac{2(a+b-2c)\pm \sqrt{4(a+b-2c{)}^2-4(a-b{)}^2}}{2}y=(a+b-2c)\pm \sqrt{(a+b-2c+a-b)(a+b-2c-a+b)}y=(a-c)+(b-c)\pm 2\sqrt{(a-c)(b-c)}=(\sqrt{(a-c)}\pm \sqrt{(b-c)}{)}^2\therefore yϵ[(\sqrt{(a-c)}+\sqrt{(b-c)}{)}^2,\infty )$

$\therefore$ Minimum value of y is $(\sqrt{(a-c)}+\sqrt{(b-c)}{)}^2$

$\therefore$ our assumption is true, $x$ is real.