The quadratic equation `2x^(2) + px + 3 = 0` has two equal roots if p = …………. .

Answer

To solve the quadratic equation \(2x^2 + px + 3 = 0\) for the value of \(p\) such that the equation has two equal roots, we can follow these steps:

### Step 1: Understand the condition for equal roots
For a quadratic equation of the form \(ax^2 + bx + c = 0\), the condition for the equation to have equal roots is that the discriminant \(D\) must be equal to zero. The discriminant is given by the formula:
\[ D = b^2 - 4ac \]
In our case, \(a = 2\), \(b = p\), and \(c = 3\).

### Step 2: Set up the discriminant equation
We can substitute the values of \(a\), \(b\), and \(c\) into the discriminant formula:
\[ D = p^2 - 4 \cdot 2 \cdot 3 \]
This simplifies to:
\[ D = p^2 - 24 \]

### Step 3: Set the discriminant equal to zero
Since we want the equation to have equal roots, we set the discriminant \(D\) to zero:
\[ p^2 - 24 = 0 \]

### Step 4: Solve for \(p\)
Now, we can solve for \(p\):
\[ p^2 = 24 \]
Taking the square root of both sides gives us:
\[ p = \pm \sqrt{24} \]
We can simplify \(\sqrt{24}\):
\[ \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} \]
Thus, we have:
\[ p = \pm 2\sqrt{6} \]

### Final Answer
The values of \(p\) for which the quadratic equation \(2x^2 + px + 3 = 0\) has two equal roots are:
\[ p = 2\sqrt{6} \quad \text{or} \quad p = -2\sqrt{6} \]

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