`(x+1)^(2)+x^(2)=0` has

Options

A. no real roots
B. one real root
C. two real root
D. four real root

Answer

To solve the equation \((x+1)^{2} + x^{2} = 0\), we will follow these steps:

### Step 1: Expand the equation
We start with the equation:
\[ (x+1)^{2} + x^{2} = 0 \]
Expanding \((x+1)^{2}\):
\[ (x+1)^{2} = x^{2} + 2x + 1 \]
Now substitute this back into the equation:
\[ x^{2} + 2x + 1 + x^{2} = 0 \]

### Step 2: Combine like terms
Now combine the \(x^{2}\) terms:
\[ 2x^{2} + 2x + 1 = 0 \]

### Step 3: Identify coefficients
This is a quadratic equation in the standard form \(ax^{2} + bx + c = 0\), where:
- \(a = 2\)
- \(b = 2\)
- \(c = 1\)

### Step 4: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by:
\[ D = b^{2} - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (2)^{2} - 4 \cdot (2) \cdot (1) = 4 - 8 = -4 \]

### Step 5: Analyze the discriminant
Since the discriminant \(D = -4\) is less than 0, this indicates that the quadratic equation has no real roots.

### Conclusion
Thus, the equation \((x+1)^{2} + x^{2} = 0\) has **no real roots**.

---