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: Write the Given Quadratic
Given equation:
(p − q)x² + (q − r)x + (r − p) = 0

Let:
a = p − q
b = q − r
c = r − p

Step 2: Check Sum of Coefficients
a + b + c = (p − q) + (q − r) + (r − p)
= 0

Since sum of coefficients is zero,
x = 1 is one root.

Step 3: Condition for Equal Roots
If roots are equal and one root is 1,
then both roots must be 1.

Product of roots = c/a

For equal roots:
c/a = 1 × 1 = 1

So,
(r − p)/(p − q) = 1

Cross multiply:
r − p = p − q

Rearranging:
r + q = 2p

Step 4: Conclusion
Thus, the quadratic has equal roots
if and only if
q + r = 2p

Final Answer:
Verified that roots are equal when q + r = 2p.