Question

if a b and c are distinct real numbers then show that the equation 1 by x minus A + 1 by x minus b + 1 by x minus 3 is equal to zero in which X is not equal to abc has real roots ​

Answer 1

Step-by-step explanation:

Given If a b and c are distinct real numbers then show that the equation 1 by x minus a + 1 by x minus b + 1 by x minus 3 is equal to zero in which x is not equal to abc has real roots

• So (x – b)(x – c) + (x – a)(x – c) + (x – b) (x – a) = 0
• So x^2 – bx – cx + bc + x^2 – ax – cx + ac + x^2 – bx – ax + ab = 0
• 3x^2 – 2(b + c + a) x + (ab + bc + ca) = 0
• D = b^2 – 4 ac
• = 4(b + c + a)^2 – 4(ab + bc + ca) (3)
• = 4[b^2 + c^2 + a^2 + 2ab + 2bc + 2ca – 3ab – 3bc – 3ca]
• = 4 (b^2 + c^2 + a^2 – ab – bc – ca)
• = 2 (2b^2 + 2c^2 + 2a^2 – 2ab – 2bc – 2ca)
• D = 2[ (b – c)^2 + (b – a)^2 + (c – a)^2]
• D = 0  if  a = b = c.

So for real roots D > 0

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https://brainly.in/question/13976103