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:

An alternative way to solve this is to substitute \(p = \frac{q + r}{2}\) directly into the discriminant formula \(D = B^2 - 4AC\) and show that it reduces to zero.
However, utilizing the root properties of equations with coefficient sums equal to zero is much faster and reduces algebra steps!

Answer 1

Step 1: Substitute the given condition into the coefficients.
Here $A = p - q$, $B = q - r$, $C = r - p$. Using $q + r = 2p$, we get $r = 2p - q$.
Step 2: Rewrite B and C in terms of p and q only.
\[ B = q - r = q - (2p - q) = 2(q - p), \qquad C = r - p = (2p - q) - p = p - q \]
So $A = p - q$, $C = p - q$ (equal to A), and $B = -2(p - q)$.
Step 3: Compute the discriminant directly.
\[ D = B^2 - 4AC = [-2(p-q)]^2 - 4(p-q)(p-q) = 4(p-q)^2 - 4(p-q)^2 = 0 \]
Since the discriminant is zero, the roots of the quadratic equation are real and equal, which is what we needed to verify.
\[ \boxed{D = 0 \implies \text{roots are equal}} \]