Show that the equation `x^(2)+2x+1+sqrtx` has no real roots.

Answer

To show that the equation \( x^2 + 2x + 1 + \sqrt{x} = 0 \) has no real roots, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 + 2x + 1 + \sqrt{x} = 0 \]
Notice that \( x^2 + 2x + 1 \) can be factored as:
\[ (x + 1)^2 \]
Thus, we can rewrite the equation as:
\[ (x + 1)^2 + \sqrt{x} = 0 \]

### Step 2: Analyze the components
Now we analyze the two components of the equation:
1. \( (x + 1)^2 \) is a perfect square and is always non-negative (i.e., \( (x + 1)^2 \geq 0 \)).
2. \( \sqrt{x} \) is also non-negative for \( x \geq 0 \) (i.e., \( \sqrt{x} \geq 0 \)).

### Step 3: Combine the inequalities
Since both \( (x + 1)^2 \) and \( \sqrt{x} \) are non-negative, their sum must also be non-negative:
\[ (x + 1)^2 + \sqrt{x} \geq 0 \]
This means that the left-hand side of our equation is always greater than or equal to zero.

### Step 4: Determine when the sum equals zero
For the equation to hold true, we need:
\[ (x + 1)^2 + \sqrt{x} = 0 \]
The only way for a sum of two non-negative numbers to equal zero is if both numbers are individually zero:
1. \( (x + 1)^2 = 0 \) implies \( x + 1 = 0 \) or \( x = -1 \).
2. \( \sqrt{x} = 0 \) implies \( x = 0 \).

### Step 5: Check the solutions
Now we have two potential solutions:
- From \( (x + 1)^2 = 0 \), we get \( x = -1 \).
- From \( \sqrt{x} = 0 \), we get \( x = 0 \).

However, \( x = -1 \) is not a valid solution because \( \sqrt{x} \) is not defined for negative values of \( x \) in the real number system. Therefore, the only viable solution from the second equation is \( x = 0 \).

### Step 6: Verify the solution
Substituting \( x = 0 \) back into the original equation:
\[ 0^2 + 2(0) + 1 + \sqrt{0} = 1 \neq 0 \]
Thus, \( x = 0 \) does not satisfy the equation.

### Conclusion
Since there are no values of \( x \) that satisfy the equation \( x^2 + 2x + 1 + \sqrt{x} = 0 \), we conclude that the equation has no real roots.

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