A(b-c)xsquare+b(c-a)x+c(a-b)=0 has its roots equal then
Answers
Answer:
With a,b,c≠0, observe that 1/a,1/b,1/c is an Arithmetic Progression iff
K=0, where
K=ab+bc−2ac.
The condition that the quadratic has a repeated root is that the discriminant is zero. We have
0=b2(c−a)2−4ac(a−b)(b−c)⟺
0=b2c2+b2a2−2b2ac−4ac(−b2+bc+ba−ac)⟺
0=b2c2+b2a2+2acb2−4ac(bc+ba−ac)⟺
0=b2c2+b2a2+2acb2−4ac(bc+ba−2ac)−4a2c2⟺
0=b2c2+b2a2+2acb2−4acK−4a2c2⟺
0=(bc+ba)2−4acK−4a2c2⟺
0=(bc+ba)2−(2ac)2−4acK⟺
0=(bc+ba−2ac)(bc+ba+2ac)−4acK
0=K(bc+ba+2ac)−4acK⟺
0=K(bc+ba−2ac)⟺
0=K2.
clearly 1 is a root of the given equation. since it is of second degree and has repeated roots the the other root is also 1. we know that sum of roots is b(a-c)/(a(b-c))=2. manipulating in proper form we will get a, b, c are in H.P
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