Question

What are roots of equation (b-c)x2+(c-a)x+(a-b)=0?

Answer 1

★ QUADRATIC RESOLUTION ★

Answer 2

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!!