If x>0, Is \sqrt{x} > x ?
(1) x is not equal to 1
(2) x\sqrt{x} >x^2
My Reasoning:
![]()
(1) x is not equal to 1
(2) x\sqrt{x} >x^2
My Reasoning:
[Reveal] Spoiler:
Request you to please post your reasoning for this. I know, putting values can be a good option, however I want to know where I am getting wrong.
1) Ofcourse, statement 1 is not sufficient.
2) x\sqrt{(x)}> x^2
x\sqrt{(x)} -x^2 > 0
Taking x common
x(\sqrt{(x)}-x)>0
it implies
eitherx>0 or\sqrt{(x)} > x
My question is how can we infer that \sqrt{(x)} > x is true.
Please explain
1) Ofcourse, statement 1 is not sufficient.
2) x\sqrt{(x)}> x^2
x\sqrt{(x)} -x^2 > 0
Taking x common
x(\sqrt{(x)}-x)>0
it implies
eitherx>0 or\sqrt{(x)} > x
My question is how can we infer that \sqrt{(x)} > x is true.
Please explain