What are roots of equation (b-c)x2+(c-a)x+(a-b)=0?
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Helloo!!
(b−c)x2+(c−a)x+(a−b)=0 ≡ ax2+bx+c=0(b−c)x2+(c−a)x+(a−b)=0 ≡ ax2+bx+c=0
Here, a =(b-c) , b = (c-a) and c = (a-b)
So, D = (c−a)2−4(b−c)(a−b)=0
c2+a2−2ac−4(ab−b2−ac+bc)=0
c2+a2−2ac−4ab+4b2+4ac−4bc=0
c2+a2+2ac+4b2−4ab−4bc=0
(c+a)2+4b2−4b(a+c)=0
(c+a)2+(2b)2−2⋅(c+a)⋅(2b)=0
[(c+a)−(2b)]2=0
c+a−2b=0
a + c = 2b
Hope it helps!!
(b−c)x2+(c−a)x+(a−b)=0 ≡ ax2+bx+c=0(b−c)x2+(c−a)x+(a−b)=0 ≡ ax2+bx+c=0
Here, a =(b-c) , b = (c-a) and c = (a-b)
So, D = (c−a)2−4(b−c)(a−b)=0
c2+a2−2ac−4(ab−b2−ac+bc)=0
c2+a2−2ac−4ab+4b2+4ac−4bc=0
c2+a2+2ac+4b2−4ab−4bc=0
(c+a)2+4b2−4b(a+c)=0
(c+a)2+(2b)2−2⋅(c+a)⋅(2b)=0
[(c+a)−(2b)]2=0
c+a−2b=0
a + c = 2b
Hope it helps!!
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