The roots of the equation `(p-q) x^(2)+(q-r) x+(r-p)=0` are

Options

A. `(p-q)/(r-p),1`
B. `(q-r)/(p-r),1`
C. `(r-p)/(p-q),1`
D. none of these

Answer

To find the roots of the quadratic equation \((p-q)x^2 + (q-r)x + (r-p) = 0\), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(ax^2 + bx + c = 0\).

### Step 1: Identify coefficients
From the equation \((p-q)x^2 + (q-r)x + (r-p) = 0\), we identify:
- \(a = p - q\)
- \(b = q - r\)
- \(c = r - p\)

### Step 2: Calculate the discriminant
The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (q - r)^2 - 4(p - q)(r - p) \]

### Step 3: Substitute into the quadratic formula
Now we substitute \(a\), \(b\), and \(D\) into the quadratic formula:
\[ x = \frac{-(q - r) \pm \sqrt{(q - r)^2 - 4(p - q)(r - p)}}{2(p - q)} \]

### Step 4: Simplify the expression
We can simplify the expression further if needed, but the roots are given by:
\[ x = \frac{r - q \pm \sqrt{(q - r)^2 - 4(p - q)(r - p)}}{2(p - q)} \]

### Final Result
Thus, the roots of the equation \((p-q)x^2 + (q-r)x + (r-p) = 0\) are:
\[ x = \frac{r - q \pm \sqrt{(q - r)^2 - 4(p - q)(r - p)}}{2(p - q)} \]