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If $\large (a+b-6)^2 + a^2+b^2+1+2b = 2ab+2a$, then the value of a is
- A7
- B6
- C3.5
- D2.5
Answer not set
Discussion 1
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CCommunity
option 3 : 3.5 is correct
$\large (a + b – 6)^2 + a^2 + b^2 + 1 + 2b = 2ab + 2a$
$\large ⇒ (a + b – 6)^2 + a^2 + b^2 +1 + 2b – 2ab – 2a = 0$
$\large ⇒ (a + b – 6)^2 + (a)^2 + (–b)^2 + (–1)2 + 2a (–b) + 2 (–b) (–1) + 2 (a) (–1) = 0$
$\large ⇒ (a + b – 6)^2 + (a – b – 1)^2 = 0$
$\large ⇒ a + b – 6 = 0$ and $\large a – b – 1 = 0$
$\large ⇒ a + b = 6 $ and $\large a – b = 1$
On adding these two equations,
$\large a + b + a – b = 6 + 1$
$\large ⇒ 2a = 7$
$\large a = 3.5$
$\large (a + b – 6)^2 + a^2 + b^2 + 1 + 2b = 2ab + 2a$
$\large ⇒ (a + b – 6)^2 + a^2 + b^2 +1 + 2b – 2ab – 2a = 0$
$\large ⇒ (a + b – 6)^2 + (a)^2 + (–b)^2 + (–1)2 + 2a (–b) + 2 (–b) (–1) + 2 (a) (–1) = 0$
$\large ⇒ (a + b – 6)^2 + (a – b – 1)^2 = 0$
$\large ⇒ a + b – 6 = 0$ and $\large a – b – 1 = 0$
$\large ⇒ a + b = 6 $ and $\large a – b = 1$
On adding these two equations,
$\large a + b + a – b = 6 + 1$
$\large ⇒ 2a = 7$
$\large a = 3.5$