The roots of the equation `2x^(2)-6x+7=0` are

Options

A. real, unequal and rational
B. real, unequal and irrational
C. real and equal
D. imaginary

Explanation

To find the roots of the quadratic equation \(2x^2 - 6x + 7 = 0\), we can follow these steps:

### Step 1: Identify coefficients
The standard form of a quadratic equation is \(Ax^2 + Bx + C = 0\). In our equation:
- \(A = 2\)
- \(B = -6\)
- \(C = 7\)

### Step 2: Calculate the discriminant
The discriminant \(D\) is given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values we identified:
\[ D = (-6)^2 - 4 \cdot 2 \cdot 7 \]
Calculating this gives:
\[ D = 36 - 56 = -20 \]

### Step 3: Analyze the discriminant
The value of the discriminant helps us determine the nature of the roots:
- If \(D > 0\), the roots are real and unequal.
- If \(D = 0\), the roots are real and equal.
- If \(D < 0\), the roots are imaginary (or complex).

Since \(D = -20 < 0\), we conclude that the roots are imaginary.

### Step 4: State the roots
Since the discriminant is negative, we can express the roots using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{D}}{2A} \]
Substituting the values:
\[ x = \frac{-(-6) \pm \sqrt{-20}}{2 \cdot 2} \]
This simplifies to:
\[ x = \frac{6 \pm \sqrt{-20}}{4} \]
We can express \(\sqrt{-20}\) as \(i\sqrt{20} = i\sqrt{4 \cdot 5} = 2i\sqrt{5}\):
\[ x = \frac{6 \pm 2i\sqrt{5}}{4} \]
This further simplifies to:
\[ x = \frac{3 \pm i\sqrt{5}}{2} \]

### Conclusion
The roots of the equation \(2x^2 - 6x + 7 = 0\) are:
\[ x = \frac{3 + i\sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - i\sqrt{5}}{2} \]

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