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# JEE Class main Answered

find discriminant
Asked by bablukrpes | 29 Sep, 2021, 10:22: PM
Expert Answer
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
x2 - (a + b)x + ab + x2 - (b + c)x + bc + x2 - (c + a)x + ac = 0
3x2 - (a + b + b + c + c + a) + ab + bc + ac = 0
3x2 - 2(a + b + c) + ab + bc + ac = 0
Discriminant = b2 - 4ac = 4(a + b + c)2 - 12(ab + bc + ac)
= 4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ac - 12ab - 12bc - 12ac
= 4(a2 + b2 + c2 - ab - bc - ac)
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