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Linear Equations in One Variable

Let a2+b2-4...

Question

Let $a^{2} + b^{2} - 4 a + 16 b + 68 = 0$ then find the value of $a + b$ .

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Solution

Now,
$a^{2} + b^{2} - 4 a + 16 b + 68 = 0$
or, $a^{2} - 4 a + 4 + b^{2} + 16 b + 64 = 0$
or, $(a - 2)^{2} + (b + 8)^{2} = 0$
Sum of two positive numbers is $0$ only when they are individually $0$ .
This given $a - 2 = 0 \Rightarrow a = 2$ and $b + 8 = 0 \Rightarrow b = - 8$ .
Then $a + b = 2 - 8 = - 6$ .

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