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

2
by 426ri543

2016-03-21T12:44:46+05:30
Given quadratic equation is (a – b)x2 + (b – c)x + (c – a) = 0

Since the root are equal, discriminent of the quadritic equation = 0

Hence,
(b – c)2 = 4(a – b)(c – a)
⇒ b2 + c2 – 2bc = 4(ac – a2 – bc + ab)
⇒ b2 + c2 – 2bc = 4ac – 4a2 – 4bc + 4ab
⇒ b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
⇒ b2 + c2 + 4a+ 2bc – 4ac – 4ab = 0
⇒ b2 + c2 + (2a)+ 2(b)(c) – 2(2a)c – 2(2a)b = 0
⇒ b2 + c2 + (–2a)+ 2(b)(c) + 2(–2a)c + 2(–2a)b = 0
⇒ (b + c – 2a)2 = 0
⇒ b + c – 2a = 0
∴ b + c = 2a

Hence Proved
• Brainly User
2016-03-21T12:47:11+05:30
Given quadratic equation is (a – b)x2 + (b – c)x + (c – a) = 0

Since the root are equal, discriminent of the quadritic equation = 0

Hence,
(b – c)2 = 4(a – b)(c – a)
⇒ b2 + c2 – 2bc = 4(ac – a2 – bc + ab)
⇒ b2 + c2 – 2bc = 4ac – 4a2 – 4bc + 4ab
⇒ b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
⇒ b2 + c2 + 4a2 + 2bc – 4ac – 4ab = 0
⇒ b2 + c2 + (2a)2 + 2(b)(c) – 2(2a)c – 2(2a)b = 0
⇒ b2 + c2 + (–2a)2 + 2(b)(c) + 2(–2a)c + 2(–2a)b = 0
⇒ (b + c – 2a)2 = 0
⇒ b + c – 2a = 0
∴ b + c = 2a

Hence Proved
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AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm. If the tangent at A and B intersect at a point P then find the length PA
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