Question

If the roots of ( a-b)×+(b-c)×+(c-a)=0 are equal, prove that 2a=b+c

Answer 1

Step-by-step explanation:

(a - b)x² + (b - c)x + (c - a)=0

Comparing with quadratic equation  with

Ax²+Bx+C=0

A=(a - b),  B=(b - c),  and C=(c - a)

We know that when roots are equal then Discriminant(D) = 0

∴ D = 0

⇒b² - 4ac = 0

⇒(b - c)² - 4(c - a)(a - b) = 0

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

⇒4a² + b² + c² - 4ab - 4ac + 2bc = 0

⇒(-2a)² + (b)² + (c)² + 2[b×(-2a) + (-2a)×c + b×c] = 0

[∵x²+y²+x²+2(xy+yz+za) = (x+y+z)²]

⇒(b - 2a + c)² = 0

⇒b - 2a + c = 0

⇒b + c = 2a

∴ 2a = b + c (Hence Proved)

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