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Given roots of the equation are equal.
Hence ∆=b²-4ac=0
Here, a=( b-c), b=( c - a), c =( a - b)
=> (c-a)²-4(b-c)(a-b) = 0
=> (c-a)²=4(b-c)(a-b)
=> c²+a²-2ac=4(ab-b²-ac+bc)
=>c²+a²-2ac=4ab-4b²-4ac+4bc
=> c²+a²-2ac-4ab+4b²+4ac-4bc=0
=> c²+a²+2ac+4b²-4bc-4ab=0
=> (c+a)²+(2b)²-4b(c+a)=0
=>(c+a)²+(2b)²-2(2b)(c+a)=0
=> [ (c+a)-2b] ²=0
=> [ (c+a)-2b] =0
=> (c+a)=2b
=> 2b=c+a
HENCE PROVED !!^^
Hence ∆=b²-4ac=0
Here, a=( b-c), b=( c - a), c =( a - b)
=> (c-a)²-4(b-c)(a-b) = 0
=> (c-a)²=4(b-c)(a-b)
=> c²+a²-2ac=4(ab-b²-ac+bc)
=>c²+a²-2ac=4ab-4b²-4ac+4bc
=> c²+a²-2ac-4ab+4b²+4ac-4bc=0
=> c²+a²+2ac+4b²-4bc-4ab=0
=> (c+a)²+(2b)²-4b(c+a)=0
=>(c+a)²+(2b)²-2(2b)(c+a)=0
=> [ (c+a)-2b] ²=0
=> [ (c+a)-2b] =0
=> (c+a)=2b
=> 2b=c+a
HENCE PROVED !!^^
apoorvkamboj798:
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