Find the zeros of the polynomial `x^(2) -3` and verify the relationship `alpha + beta = - (b)/(a)` if `alpha, beta` are its zeros.

Answer

To find the zeros of the polynomial \(x^2 - 3\) and verify the relationship \(\alpha + \beta = -\frac{b}{a}\), where \(\alpha\) and \(\beta\) are the zeros, follow these steps:

### Step 1: Write the polynomial in standard form
The polynomial is already in standard form:
\[ x^2 - 3 = 0 \]

### Step 2: Identify coefficients
From the polynomial \(x^2 - 3\), we can identify:
- \(a = 1\) (coefficient of \(x^2\))
- \(b = 0\) (coefficient of \(x\))
- \(c = -3\) (constant term)

### Step 3: Use the quadratic formula
The quadratic formula to find the roots (zeros) of a polynomial \(ax^2 + bx + c = 0\) is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ x = \frac{-0 \pm \sqrt{0^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]

### Step 4: Simplify under the square root
Calculating the discriminant:
\[ b^2 - 4ac = 0 - 4 \cdot 1 \cdot (-3) = 12 \]
Now substituting back into the formula:
\[ x = \frac{0 \pm \sqrt{12}}{2} \]

### Step 5: Simplify the square root
\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]
Thus, we have:
\[ x = \frac{0 \pm 2\sqrt{3}}{2} = \pm \sqrt{3} \]

### Step 6: Identify the zeros
The zeros of the polynomial \(x^2 - 3\) are:
\[ \alpha = \sqrt{3}, \quad \beta = -\sqrt{3} \]

### Step 7: Verify the relationship \(\alpha + \beta = -\frac{b}{a}\)
Now, we calculate \(\alpha + \beta\):
\[ \alpha + \beta = \sqrt{3} + (-\sqrt{3}) = 0 \]

Next, we calculate \(-\frac{b}{a}\):
\[ -\frac{b}{a} = -\frac{0}{1} = 0 \]

### Step 8: Conclusion
Since \(\alpha + \beta = 0\) and \(-\frac{b}{a} = 0\), we have verified that:
\[ \alpha + \beta = -\frac{b}{a} \]
Thus, the relationship is satisfied.