Question

Verify that roots of the quadratic equation \((p - q)x^2 + (q - r)x + (r - p) = 0\) are equal when \(q + r = 2p\).

Hint:

If you see a quadratic equation where the coefficients are cyclic (like \(a-b, b-c, c-a\)), always check if their sum is zero. If it is, \(x=1\) is a guaranteed root!

Answer 1

Step 1: Understanding the Concept:
For a quadratic equation \(ax^2 + bx + c = 0\) to have equal roots, the discriminant \(D = b^2 - 4ac\) must be zero.
Alternatively, if the sum of coefficients is zero, then \(x=1\) is one of the roots.
Step 2: Key Formula or Approach:
Sum of coefficients: \((p - q) + (q - r) + (r - p) = 0\).
If \(x=1\) is a root and roots are equal, both roots must be 1.
In that case, the product of roots \(\frac{c}{a} = 1 \cdot 1 = 1\).
Step 3: Detailed Explanation:
Let the coefficients be \(A = p-q\), \(B = q-r\), and \(C = r-p\).
Notice that \(A + B + C = (p - q) + (q - r) + (r - p) = 0\).
This means \(x = 1\) is always a root of this equation.
For the roots to be equal, both roots must be \(x = 1\).
We know the product of roots is \(\frac{C}{A}\).
So, \(\frac{r - p}{p - q} = 1\).
\[ r - p = p - q \]
Rearranging the terms:
\[ r + q = p + p \]
\[ q + r = 2p \]
This condition matches the given condition \(q + r = 2p\).
Alternatively, checking the discriminant:
\[ D = (q - r)^2 - 4(p - q)(r - p) \]
Substitute \(p = \frac{q + r}{2}\) into the expression and solve for \(D = 0\).
The property of cyclic coefficients (\(A+B+C=0\)) is a more elegant way to verify.
Step 4: Final Answer:
Thus, it is verified that the roots are equal when \(q + r = 2p\).