34. If the roots of the quadratic equation (x – a) (x – b) + (x – b)(x – c) + (x - C)(x – a) = 0 are equal. Then,
show that a = b = c.
Answers
Step-by-step explanation:
(x – a) (x – b) + (x – b)(x – c) + (x – c)(x – a) = 0
x(x – b) – a(x – b) + x(x – c) – b(x – c) + x(x – a) – c(x – a) = 0
x² – xb – ax + ab + x² – xc – xb + bc + x² – ax – xc + ac = 0
3x² – xb – ax – xc – xb – ax – xc + ab + bc + ac = 0
3x² – 2xb – 2ax – 2xc + ab + bc + ac = 0
3x² – 2(xb + ax + xc) + (ab + bc + ac) = 0
3x² – 2(a + b + c)x + (ab + bc + ac) = 0
- x² coefficient = 3
- x coefficient = –2(a + b + c)
- constant = ab + bc + ac
This is in the form of the quadratic equation.
Nature of roots is determined by the values of the discriminant.
Discriminant, D = 0
(coefficient of x)² – 4(x² coefficient)(constant) = 0
[–2(a+b+c)]² – 4(3)(ab + bc + ac) = 0
4(a + b + c)² – 12(ab + bc + ac) = 0
4{a² + b² + c² + 2(ab + bc + ac)} – 12(ab + bc + ac) = 0
4a² + 4b² + 4c² + 8ab + 8bc + 8ac – 12ab – 12bc – 12ac = 0
4a² + 4b² + 4c² – 4ab – 4bc – 4ac = 0
4(a² + b² + c² – ab – bc – ac) = 0
a² + b² + c² – ab – bc – ac = 0/4
a² + b² + c² – ab – bc – ac = 0
Multiply 2 on both sides,
2(a² + b² + c² – ab – bc – ac) = 0
2a² + 2b² + 2c² – 2ab – 2bc – 2ac = 0
a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ac = 0
a² + b² – 2ab + b² + c² – 2bc + a² + c² – 2ac = 0
(a – b)² + (b – c)² + (c – a)² = 0
This is possible only when a = b = c
Hence proved!
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Identities used :
- (x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
- x² + y² – 2xy = (x – y)²
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