Question

If the roots of the quadratic equation (p - q)x2 + (q-r)x + (r-p) = 0 are equal, then show
that 2p = q +r.​

Answer 1

Answer:

When roots are real:

D = 0

D = b^2 - 4ac = 0

=> (q - r)^2 - 4(r - p)(p - q) = 0

=> q^2 + r^2 - 2qr - 4rp + 4qr + 4p^2 - 4pq = 0

=> q^2 + r^2 + 4p^2 + 2qr - 4rq - 4pq = 0

=> (q + r - 2p)^2 = 0

=> q + r - 2p = 0

=> 2p = q + r

proved.

Identity used: (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

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