The number of real roots of the equation `2x^(4) + 5x^(2) + 3 = 0`, is

Options

A. 4
B. 1
C. 0
D. 3

Answer

To find the number of real roots of the equation \(2x^4 + 5x^2 + 3 = 0\), we can follow these steps:

### Step 1: Substitute \(x^2\) with \(k\)
Let \(k = x^2\). Then the equation becomes:
\[ 2k^2 + 5k + 3 = 0 \]

### Step 2: Identify the coefficients
In the quadratic equation \(2k^2 + 5k + 3 = 0\), we have:
- \(a = 2\)
- \(b = 5\)
- \(c = 3\)

### Step 3: Calculate the discriminant
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = 5^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \]

### Step 4: Analyze the discriminant
Since the discriminant \(D = 1\) is greater than 0, the quadratic equation \(2k^2 + 5k + 3 = 0\) has 2 distinct real roots.

### Step 5: Find the roots for \(k\)
Using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ k = \frac{-5 \pm \sqrt{1}}{2 \cdot 2} = \frac{-5 \pm 1}{4} \]
Calculating the roots:
1. \(k_1 = \frac{-5 + 1}{4} = \frac{-4}{4} = -1\)
2. \(k_2 = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2}\)

### Step 6: Analyze the roots for \(x^2\)
Since \(k = x^2\), we have:
- \(x^2 = -1\)
- \(x^2 = -\frac{3}{2}\)

Both values for \(k\) are negative.

### Step 7: Conclusion on real roots
Since \(x^2\) cannot be negative for real numbers, there are no real solutions for \(x\).

Thus, the number of real roots of the equation \(2x^4 + 5x^2 + 3 = 0\) is **0**.

### Final Answer:
The number of real roots is **0**.

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