For a complex number z, the equation `z^(2)+(p+iq)z+ r+" is "=0` has a real root (where p, q, r, s are non - zero real numbers and `i^(2)-1`), then

Options

A. `pqr=r^(2)+p^(2)s`
B. `prs=q^(2)+r^(2)p`
C. `prs=p^(2)+s^(2)q`
D. `pqs=s^(2)+q^(2)r`

Answer

To solve the given problem, we need to analyze the quadratic equation in the complex number \( z \):

\[ z^2 + (p + iq)z + (r + is) = 0 \]

where \( p, q, r, s \) are non-zero real numbers and \( i^2 = -1 \). We are tasked with finding conditions under which this equation has a real root.

### Step 1: Assume a Real Root
Let \( z = \alpha \), where \( \alpha \) is a real number. Substituting \( \alpha \) into the equation gives:

\[ \alpha^2 + (p + iq)\alpha + (r + is) = 0 \]

### Step 2: Separate Real and Imaginary Parts
Expanding the equation, we have:

\[ \alpha^2 + p\alpha + iq\alpha + r + is = 0 \]

We can separate this into real and imaginary parts:

- Real part: \( \alpha^2 + p\alpha + r = 0 \)
- Imaginary part: \( q\alpha + s = 0 \)

### Step 3: Solve the Imaginary Part
From the imaginary part, we can solve for \( \alpha \):

\[ q\alpha + s = 0 \implies \alpha = -\frac{s}{q} \]

### Step 4: Substitute \(\alpha\) into the Real Part
Now, substitute \( \alpha = -\frac{s}{q} \) into the real part:

\[ \left(-\frac{s}{q}\right)^2 + p\left(-\frac{s}{q}\right) + r = 0 \]

This simplifies to:

\[ \frac{s^2}{q^2} - \frac{ps}{q} + r = 0 \]

### Step 5: Clear the Denominator
To eliminate the fraction, multiply through by \( q^2 \):

\[ s^2 - psq + rq^2 = 0 \]

### Step 6: Rearrange the Equation
Rearranging gives:

\[ s^2 + rq^2 = psq \]

### Step 7: Final Form
This can be rearranged to match the form of one of the options:

\[ psq = s^2 + rq^2 \]

Thus, we find that the condition for the quadratic equation to have a real root is:

\[ p \cdot q \cdot s = s^2 + r \cdot q^2 \]

This matches with option **(d)**.

### Conclusion
The correct answer is option **(d)**: \( p \cdot q \cdot s = s^2 + q^2 \cdot r \).

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