Question : If $(a+b-6)^{2}+a^{2}+b^{2}+1+2b=2ab+2a$, the value of $a$ is:
Option 1: 7
Option 2: 6
Option 3: 3.5
Option 4: 2.5
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Correct Answer: 3.5
Solution :
Given:
$(a+b-6)^{2} + a^{2}+b^{2}+1+2b = 2ab + 2a$
⇒ $(a+b-6)^{2}+a^{2}+ b^{2}+1+2b-2ab-2a = 0$
⇒ $(a+b-6)^{2}+ a^{2}+ (-b)^{2}+ (-1)^{2}+2a (-b) +2 (-b) (-1)+2 (a) (-1)=0$
⇒ $ (a + b-6)^{2}+(a-b-1)^{2}=0$
⇒ $a+b-6 = 0$ and $a-b-1=0$ [if the sum of the squares of two numbers are zero then each of them will also be zero]
⇒ $a+b = 6$ and $a-b = 1$
On adding these two equations,
$a+b+a-b = 6+1$
⇒ $2a = 7$
⇒ $a=\frac{7}{2}=3.5$
Hence, the correct answer is $3.5$.
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