Question

the root of quadratic Polynomial (b-c)x2+(c-a)x+(a-b)=0 are real show that b2-4ac=0
pls fast​

Answer 1

Answer:

0

Step-by-step explanation:

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!

I hope it helps!