Question

If the roots of the equation(b-c)x²+(c-a)x+(a-b)=0 are equal, then prove that 2b = a + c.

Answer 1

Step-by-step explanation:

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Answer 2

2b = a + c  is proved

Step-by-step explanation:

Roots of (b-c)x²+(c-a)x+(a-b)=0 are equal.

So discriminant is zero.

d = b² - 4ac

a = b-c

b = c-a

c = a-b

Substituting to find d:

(c-a)² - 4(b-c)(a-b) = 0

c² + a² -2ac -4 (ab -b² + bc - ac) = 0

c² + a² -2ac -4ab +4b² -4bc +4ac = 0  

a² + (-2b)² + 2a(-2b) + +2(-2b)c +2ac = 0

(a -2b + c)² = 0 from algebric expression of (x + y + z)² = x² + y² + z² + 2xy + 2yz +2az

a - 2b + c = 0

2b = a + c

Hence proved.