Square root greater than its Square? Impossible!

We know that square of any value b/w 0<x<1, then its resultant value less than its original value. apply square both side, here only one value 1/4

$\sqrt{x}>x^2,~~~~~squaring~ both~ sides,~~x>x^4 ~is~the~condition.\\ For~-1<x<1:-~~~~~~if~~ n>1, ~~x^n<x.\\ n=4>1.~~So~we~should~have,~number~|x|<1.\\ 1/4~~is ~the~only~such~number~from~the~list.\\ Answer=\color{#D61F06}{1/4}.$

I chose $\frac{1}{4}$ , because:

$\square{\frac{1}{4}}=\frac{1}{2}$

so $\frac{1}{2}>\frac{1}{4}$

root of 1/4 is 1/2, that is 0.5.. square of 1/4 is 1/16, that is 0.0625.. therefore 0.5 is greater than 0.0625

The numerator should be minimum as possible, it can't be zero since square root of 0 is equal to square of zero . Thus it should be one , only one option contains it .

Check for the given options as per given condition

If $a<b$ then $(\frac{a}{b})^2=\frac{a^2}{b^2}$ which is smaller because $b-a<b^2-a^2 \implies b-a<(b-a)(b+a)$

It simply telling that x<1... So just option checking..😉

Let x=1/4, and 1/4=0.25 and its(1/4) square root is 0.5 while the square of 1/4 =0.0625. Thus, x = 1/4 in that the square root of 1/4 which is 0.5 is > then (1/4)^2 which is 0.0625.

1/4=1/2 so 1/2=1/4

Yah ! 3000 solvers.

$\sqrt{x} > x^2 => x > x^4$

Thus for any value of $x$ for which $-1 < x < 1$ excluding $0$ , the multiple of itself will yield a value less than itself (i.e 0.5 x 0.5 = 0.25 < 0.5)

So hence, $\boxed{x = \frac{1}{4}}$

because the square of 0<x<1 is smaller always than their respective square roots :)

if any no is less then one the its square is always less then square root of it self

If x = 1/4 then 1/4 ^ 1/2 is equal to 1/2 which is equal to the square root of x

x^1/2 > x ; 1/2 >1/4

1\4 is than 1\2

let x=1/4 then square root of 1/4 is 1/2 so 1/2=.5 is greater than 1/4=.25. so ans is 1/4. best wishes next

sqrt(1/4) = 1/2, half is greater than 1/4

Write a solution.

sqrt(1/4)=0.5 >(1/4) ((0.5)^2) basically given any real number n, such that 0<n<1 will satisfy the given condition for further explanation try graphing both y=sqrt(x) and y=x on the same axis

which is greater value cos6 or cos60

1/4 = 0,25. √0,25 = 0,5. This satisfies the proposition.

$\sqrt{x} > x$ will happen when $0 < x < 1$

Then, the answer is $\frac{1}{4}$ .