Question

if p,q and r are the real numbers then roots of the eqn (x-p)(x-q)+(x-q)(x-r)+(x-p)(x-r)=0

Answer 1

p, q and r are the real numbers then the roots of the equation (x - p)(x - q) + (x - q)(x - r) + (x - r)(x - p) = 0 are ...

solution : let's rearrange the equation into simpler form.

(x - p)(x - q) + (x - q)(x - r) + (x - r)(x - p) = 0

⇒{x² - (p + q)x + pq} + {x² - (q + r)x + qr} + {x² - (p + r)x + pr} = 0

⇒3x² -{(p + q) + (q + r) + (r + p)}x + (pq + qr + rp) = 0

⇒3x² - 2(p + q + r)x + ( pq + qr + rp ) = 0

Discriminant, D = b² - 4ac

= {2(p + q + r)}² -4(pq + qr + rp)(3)

= 4[p² + q² + r² - pq - qr - rp] ≠ 0

so, using formula,

x = {- b ± √D}/2a

so, x = {2(p + q + r) ± 2√(p² + q² + r² - pq - qr - rp)}/2×3

= 1/3[(p + q + r) ± √(p² + q² + r² - pq - qr - rp) ]

Therefore the roots of equation are ; 1/3[(p + q + r) ± √(p² + q² + r² - pq - qr - rp) ]

Answer 2

Answer:

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