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Mathematics 17 Online
iuytyuioiuytyuiop:

Given the function g(d)=d−15−−−−−√4, solve for g(d)=2. Give an exact answer; do not round.

iuytyuioiuytyuiop:

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AZ:

okay, this isn't too hard but it depends on how much you know so we're saying g(d) think of g(d) as your f(x) or as y and basically instead of x, they're using the letter 'd' okay So they want to know when y is 2, what is x? That is, what is 'd'?

AZ:

so we have to solve this for d \(\Large 2 = \sqrt[4]{d-15}\)

AZ:

Do you know how to get rid of that root? Be careful- It's not a square root. It has that little 4 there If it was a square root, it would be a 2 or sometimes they don't even write the 2 because it's assumed to be a 2

iuytyuioiuytyuiop:

no i dont kno ho to get rid of it

AZ:

Okay, that's fine. What would you do if you had square root? How would you get rid of a square root would you square it on both sides that is, everything to the power of 2? for example 10 = \(\sqrt{x}\) so then you do \( 10^2 = (\sqrt{x})^2\) to get \( 10^2 = x\) \( 100 = x\) does that make sense?

iuytyuioiuytyuiop:

yep

AZ:

so since we have \(\sqrt[4]{d-15}\) to get rid of that 4th root we need to take everything to the power of 4 on both sides does that make sense?

iuytyuioiuytyuiop:

yep

AZ:

so we go from \(\Large 2 = \sqrt[4]{d-15}\) to \(\Large (2)^4 = (\sqrt[4]{d-15})^4\) and end up with \((2)^4 = d-15\)

AZ:

do you know what 2^4 is

iuytyuioiuytyuiop:

16

AZ:

good so now we have 16 = d - 15 can you solve for `d` ?

iuytyuioiuytyuiop:

d = 31?

AZ:

You got it!

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