How do I decide if it’s a real solutions or extraneous solutions?

Mathematics

1 answer:

olga55 [171]2 years ago

3 0

Step-by-step explanation:

So you can check for extraneous solutions by plugging in the 5 and -2 as x.

Plug 5 in as x

$5=\sqrt{3(5)+10}\\5=\sqrt{15+10}\\5=\sqrt{25}\\5=5$

I'm not sure if this is exactly considered an extraneous solution, but I'm assuming it is, since technically the radical doesn't have a plus/minus in front of it, so it's the principle square root which means it only outputs the positive values. This would mean that the -2 wouldn't be a solution, since the principle square root doesn't output negative values, so it's a function, since if it outputted 2 values, then it wouldn't be a function by definition.

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