Algebra
- If a2 + b2 + c2 = ab + bc + ca, where a, b, c are non zero real numbers,
then the value of a + b is c
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a2 + b2 + c2 = ab + bc + ca
⇒ 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
⇒ a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ac + a2 = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
∴ a – b = 0 ⇒ a = b
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
∴ a = b = c∴ a + b = a + a = 2 c a Correct Option: A
a2 + b2 + c2 = ab + bc + ca
⇒ 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
⇒ a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ac + a2 = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
∴ a – b = 0 ⇒ a = b
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
∴ a = b = c∴ a + b = a + a = 2 c a
- If a = x + y, b = x – y, c = x + 2y, then a2 + b2 + c2 – ab – bc – ca is
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a – b = x + y – x + y = 2y
b – c = x – y – x – 2y = – 3y
c – a = x + 2y – x – y = y
Now,
a2 + b2 + c2 – ab – bc – ca= 1 (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) 2 = 1 [(a − b)2 + (b − c)2 + (c − a)2] 2 = 1 [(2y)2 + (−3y)2 + y2] 2 = 1 × 14y2 = 7y2 2 Correct Option: D
a – b = x + y – x + y = 2y
b – c = x – y – x – 2y = – 3y
c – a = x + 2y – x – y = y
Now,
a2 + b2 + c2 – ab – bc – ca= 1 (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) 2 = 1 [(a − b)2 + (b − c)2 + (c − a)2] 2 = 1 [(2y)2 + (−3y)2 + y2] 2 = 1 × 14y2 = 7y2 2
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If the expression x2 + tx + y2 is a perfect square, then the values of t is y2 4
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(a ± b)2 = a2 ± 2ab + b2
If a = x ; b = y y 2
then,± 2ab = ± 2 × x × y ± x y 2
∴ tx = ± x
⇒ t = ± 1Correct Option: A
(a ± b)2 = a2 ± 2ab + b2
If a = x ; b = y y 2
then,± 2ab = ± 2 × x × y ± x y 2
∴ tx = ± x
⇒ t = ± 1
- If x – y = 2, xy = 24, then the value of (x2 + y2) is :
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x2 + y2 = (x – y)2 + 2xy
= 4 + 2 × 24 = 52Correct Option: D
x2 + y2 = (x – y)2 + 2xy
= 4 + 2 × 24 = 52
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For real a, b, c if a2 + b2 + c2 = ab + bc + ca, then value of a + c is b
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a2 + b2 + c2 = ab + bc + ca
⇒ 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
⇒ a2 – 2ab + b2 + b2 – 2bc + c2 +
c2 – 2ac + a2 = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
⇒ a – b = 0 ⇒ a = b
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
⇒ a = b = c∴ a + c = a + a = 2 b a Correct Option: B
a2 + b2 + c2 = ab + bc + ca
⇒ 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
⇒ a2 – 2ab + b2 + b2 – 2bc + c2 +
c2 – 2ac + a2 = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
⇒ a – b = 0 ⇒ a = b
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
⇒ a = b = c∴ a + c = a + a = 2 b a
