Question

Find the values of y for which the following will be positive, negative or zero. $y=x-6\sqrt{x}+8$

• A. $y>0\forall xϵ(-\infty ,2)\cup (4,\infty )y<0\forall xϵ(2,4)y=0\forall xϵ\{2,4\}$
• B. $y>0\forall xϵ(-\infty ,4)\cup (16,\infty )y<0\forall xϵ(4,16)y=0\forall xϵ\{4,16\}$
• C. $y>0\forall xϵR$
• D. $y>0\forall xϵ(0,4)\cup (16,\infty )y<0\forall xϵ(4,16)y=0\forall xϵ\{4,16\}$

Answer 1

The correct option is D $y>0\forall xϵ(0,4)\cup (16,\infty )y<0\forall xϵ(4,16)y=0\forall xϵ\{4,16\}$

Here $y=x-6\sqrt{x}+8$; let $\sqrt{x}=a;\Rightarrow y=a^2-6a+8$
Solving the quadratic in a, we get
a=2 or 4 $\Rightarrow$ x=4 or 16
Also the co-efficient of $a^2$ is 1 i.e. positive,
$\Rightarrow$ y > 0 $\forall$ a < 2 & a > 4
$\Rightarrow$ y > 0 $\forall$ x < 4 & x > 16
And, y < 0 $\forall$ 2<a<4
$\Rightarrow$ y < 0 $\forall$ 4 < x < 16
However we assumed $a=\sqrt{x}$; notice that x can never be negative as $\sqrt{x}$ is not defined for that. Hence the values of a get restricted. The domain of this equation is not Real Number, it’s all non-negative numbers. Therefore
$y>0\forall xϵ(0,4)\cup (16,\infty )$
$y<0\forall xϵ(4,16)$
$y=0\forall xϵ\{4,16\}$