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14 Jun 2022, 10:06
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Hi!
I understand that if you are told X=sqroot(16) that you only take the positive value, just positive 4.
However, I came across a problem from Manhattan Prep below:
If g(x)=3x+root(x), what is the value of g(d^2 + 6d +9)
Manhattan Prep then showed:
3d^2+18d+27+(d+3)
AND
3d^2+18d+27-(d+3)
as answers.
If the square root is already in the problem, why do you also need the negative version of (d+3)?
Thank you!
I understand that if you are told X=sqroot(16) that you only take the positive value, just positive 4.
However, I came across a problem from Manhattan Prep below:
If g(x)=3x+root(x), what is the value of g(d^2 + 6d +9)
Manhattan Prep then showed:
3d^2+18d+27+(d+3)
AND
3d^2+18d+27-(d+3)
as answers.
If the square root is already in the problem, why do you also need the negative version of (d+3)?
Thank you!
24 Jun 2022, 10:28
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Bookmarks
woohoo921
Hi!
I understand that if you are told X=sqroot(16) that you only take the positive value, just positive 4.
However, I came across a problem from Manhattan Prep below:
If g(x)=3x+root(x), what is the value of g(d^2 + 6d +9)
Manhattan Prep then showed:
3d^2+18d+27+(d+3)
AND
3d^2+18d+27-(d+3)
as answers.
If the square root is already in the problem, why do you also need the negative version of (d+3)?
Thank you!
I understand that if you are told X=sqroot(16) that you only take the positive value, just positive 4.
However, I came across a problem from Manhattan Prep below:
If g(x)=3x+root(x), what is the value of g(d^2 + 6d +9)
Manhattan Prep then showed:
3d^2+18d+27+(d+3)
AND
3d^2+18d+27-(d+3)
as answers.
If the square root is already in the problem, why do you also need the negative version of (d+3)?
Thank you!
First it's important to make sure we know where the (d+3) came from. d^2 + 6d + 9 = (d+3)^2
Something to note about (d+3)^2 is that it is never *negative*. It is 0 if d = -3, and everywhere else, it is positive.
But (d+3) *could* be negative.
So when you take the square root of (d+3)^2, we only want to deal with 'positive' cases. In fact, it's a rule worth remembering that √(x^2) = |x|.
It only equal x if you KNOW x is positive. If x < 0, then |x| = -x
So √(d+3)^2 = (d+3) if d+3>0.
If d+3 < 0, then √(d+3)^2 = -(d+3)
(Try some simple numbers to verify this. Try d = 2 and d = -5
d=2
√5^2 = 5, = d+3
d = -5
√(-2)^2 = 2, = -(d+3)
So since we don't know if d+3 > 0 or < 0, when we write the solution, we have to consider both cases.
You are right that when we are just given the square root, we don't have to worry about +/-, we only have to worry about the positive outcome...
But here, we don't know which *is* the positive outcome, (d+3) or -(d+3)
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