If $a$ and $b$ are numbers such that $(a - 4) (b + 6) = 0$ , then what is the smallest possible value of $a^{2} + b^{2}$ ?

16

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52

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36

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10

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20

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Solution

The correct option is A $16$
Since, $(a - 4) (b + 6) = 0$
So, the possible solutions are $a = 4$
If $b$ is any value and $b$ $= - 6$
To get the smallest possible value of $a^{2} + b^{2}$ , we have to choose either $a$ or $b$ needs to be $0$ .
i.e $0^{2} + (- 6)^{2} = 36$
In this case, we have $b = - 6$ this may lead to the more value of $a^{2} + b^{2}$ if $a$ $= 0$ .
While, if we choose $b = 0$ this may lead to the smallest value of $a^{2} + b^{2} = 4^{2} + 0^{2} = 16$
Hence, option $A$ is correct.

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