Question

If the roots of equations (b-c) xsquare +(c-a) x+(a-b) =0then they are equal prove that2b=a+c

Answer 1

Step-by-step explanation:

(b - c)x² + (c - a)x + (a - b)=0

Comparing with quadratic equation  with

Ax²+Bx+C=0

A=(b - c),  B=(c - a),  and C=(a - b)

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

∴ D = 0

⇒b² - 4ac =0

⇒(c - a)² - 4(b - c)(a - b)=0

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

⇒a² + 4b² + c² - 4ab - 4bc + 2ac = 0

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

[∵x²+y²+x²+2(xy+yz+za) = (x+y+z)²]

⇒(a - 2b + c)² = 0

⇒a - 2b + c = 0

⇒a + c = 2b

∴ 2b = a + c (Hence Proved)

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