Question

(3a + 1)^2 + (b -1)^2 + (2c - 3)^2 = 0. then 3a + b + 2c = ?
say the answer step by step
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Answer 1

Answer:

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We know that if the sum of squares of three Real Numbers are zero then they are separately zero.

It can be written as

If ${x}^{2} + {y}^{2} + {z}^{2} = 0$

Then, $x = 0 \: \: , \: \: y = 0 \: \: , \: \: z = 0$

GIVEN

$(3a + 1)^2 + (b -1)^2 + (2c - 3)^2 = 0$

TO DETERMINE

$3a + b + 2c$

CALCULATION

$(3a + 1)^2 + (b -1)^2 + (2c - 3)^2 = 0$

Now by the above mentioned formula if the sum of squares of three Real Numbers are zero then they are separately zero we get

$3a + 1 = 0 \: \: , \: \: b - 1 = 0 \: \: , 2c - 3 = 0$

Which gives

$3a = - 1 \: \: , \: \: b = 1 \: \: , \: \: 2c = 3$

Hence

$3a + b + 2c = - 1 + 1 + 3 = 3$