Question

Verify that roots of the quadratic equation (p – q)x² + (q – r)x + (r – p) = 0 are equal when q + r = 2p.

Answer 1

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 `C/A`.

So, `(r - p)/(p - q) = 1`.

r – p = p – q

r + q = p + p

q + r = 2p

This condition matches the given condition q + r = 2p.

For a quadratic equation ax² + bx + c = 0 to have equal roots, the discriminant D = b² – 4ac must be zero.

Discriminant (D) = (q – r)² – 4(p – q)(r – p)

Substitute `p = (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.

Thus, it is verified that the roots are equal when q + r = 2p.