Practice your radicals!

Squaring both sides and then rearranging the above equation we get

$x^{2} -2x + 1$ =0 $(x-1)(x-1)$ =0 or x=1

But after substituting the value in equation we get

1=-1 which is not possible hence the above equation has no solution.

Since we're trying to find real solutions for $x$ , it follows from LHS of the given equation (by domain of real-valued square root function) that,

$2x-1\geq 0\implies x\geq \frac{1}{2}~\ldots~(i)$

Also, since the square root function returns non-negative value for real $x$ in domain, it follows that RHS $\geq 0$ , i.e.,

$(-x)\geq 0\implies x\leq 0~\ldots~(ii)$

If a real $x$ is a solution, then it must simultaneously satisfy the inequalities $(i)$ and $(ii).$ But note that,

$x\leq 0~\land~x\geq \frac{1}{2}\implies x\in\emptyset$

Hence, No real solution for $x$ exists .

Note: $\emptyset$ denotes empty set.

The advantage of this method is that you don't need to solve the equation itself. You just need to examine the domain and range of the function in LHS.

If $x$ is positive, the RHS is negative, which cannot be since the range of $\sqrt{\cdot}$ is positive.

If $x$ is nonpositive, the LHS is imaginary, which again cannot be since $-x$ is real if $x$ is real.

From the domain , we find that :

x is greater than or equal 1/2

i.e

x is positive

So

The right hand side of the given equation is negative whereas the left one is positive.

So

No solution exist

$\sqrt{2x-1} = -x$

Square on both sides:

$\Rightarrow (\sqrt{2x -1})^2 = (-x)^2$

$\Rightarrow 2x-1 = x^2$

So $x =1$ .

Verifying solution:

$\sqrt{2\times 1 - 1} \ne -1$

No solution exists.

A square root has only positive values, otherwise it wouldn't be a function. For those asking for x^2 = 4, the reason why that equation has 2 roots is because you don't use sqrt to solve, you have to do it like this:

x^2 = 4

x^2 - 4 =0

(x - 2)(x + 2) = 0

Hence the 2 solutions.

Good question! I made this mistake on a math test once and never made it again! :D I thought that since a root of a square had two solutions this did too, but obviously not. The only thing you need to know is that the root has only one soltion, so you cannot have a real solution of one.

SQUARE ROOT OF A NUMBER IS NEVER NEGATIVE

The simplest explanation: No real square root can be negative, so this equation should have no solution.

Root-over something is either 0 or positive number. So 'NO SOLUTION EXISTS' is the correct answer.

Square root of any number can't be negative except complex number.

In order for the root to be real, 2x > 1. Or x > 1/2. This means -x must be negative which contradicts the principle square root (which must be positive) on the left side of the equation.

as root never gives a negative number so the x must be -ve (result of root +ve) and after solving we get +ve x which contradict the domain of x so there is no solution for this equation.

Positive Square root of any real number cannot be negative...

If we square both sides of the equation, we get:

2x -1 = (x^2)

(x^2) - 2x + 1 = 0

the LHS can be re-written as:

((x - 1)^2) = 0

Which gives x=1

This implies that:

Square root of 1 is equal to negative 1

but we know that sqrt(1) is also equal to 1

so, sqrt(1) = -1

sqrt(1) =1

this means 1 = -1

Clearly that is impossible.

So, we can conclude that there is no solution

To find real solution for x:

1. Quantity inside square root should always be positive. Which means, 2x-1 >= 0 or x >= 1/2

Analyze quantity on RHS of given equation: If x>1/2, then quantity on right hand side of the given equation is always positive.

Analyze quantity on LHS of given equation: if x>=1/2, the quantity on LHS will become negative.

Positive quantity can never be equal to a negative quantity, so no real solution exists.

If you square both sides to solve for x, you should remember the fact that [sqrt(x)]^2= |x| , not just x.

Although, for this problem, you know the domain of reals is restricted to (.5,inf).

When you solve for |2x-1|, you solve for 2x-1=x^2 (x-1)(x-1) = 0 | x=1, but clearly it isn't a valid solution because sqrt(1) =/= -1 . If sqrt(1) did in fact equal -1, then that would mean that 1^(1/2) could be a negative value.

It is true that for any positive integers a,b: a^b cannot be negative.

Solving 1-2x = x^2 leaves you with x= -1+sqrt(2) and -1-sqrt(2) (two extraneous solutions).

Neither of these values are larger than 1/2, therefore, there are 0 real solutions to the problem.

from left side 2x-1 ≥0 so x ≥1/2 1st condition from right side -x ≥0 so x ≤ 0 2nd condition from both these condition we can conclude there is no real solution.

Square root only gives positive value, 'x' must be negative. Cross check the only two negative options, i.e. -2 and -1 in L.H.S. of equation. You ll get a complex number. Hence, no solution.

Square root of any number, can't be negative.. So it has no solution..As simple as that. :)

I think that this is already obvious that root square won't be ngeative right? In exception of imaginer number which is rootsquare of -1 haha

simple.....for y to be real.... y^1/2 cannot b negative and here we cannot put any negative value of x since sq. root of negative is unreal

the square root of any number has to be positive so -x doesn't work.

Simply enough, no real value can give a negative output, and by using the condition for a real number, 2x-1>0 or 0 If x = 1/2 , the bare minimum for the equality to hold, then the equation does not hold. However, for any values larger than 1/2, the square root of 2x-1 is ALWAYS positive. As any value larger than 1/2 is positive, it is safe to simply say that no real solution exists.

simply see that LHS is defined for x>=1/2 and RHS being output of square root(eventh root) must be non-negative=> -x>=0=>x<=0 .Upon intersection no common 'x' is possible.