(b-c)x2 +(c-a)x+(a-b) =0 then show 2b=a+c
Answers
ans is proved check the pic
(b-c)x²+(c-a)x+(a-b)=0
Comparing with quadratic equation
Ax²+Bx+C=0
A=(b-c),B=c-a,C=a-b
Discriminate when roots are equal
D=B²-4AC=0
D=(c-a)²−4(b-c)(a-b)=0
D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0
D=c²+a²−2ac-4ab+4ac+4b²-4bc=0
c²+a²+2ac-4b(a+c)+4b²=0
(a+c)²-4b(a+c)+4b²=0
[(a+c)-2b]²=0
a+c=2b
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Thanks for A2a!!
If roots of a quadratic equation are equal, then the discriminant of the quadratic equation is 0.
D=b2−4ac=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
Hence proved...