Can't get this one started: 4 sqrt[x] = 2x + k

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masters

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\(\displaystyle 4\cdot \sqrt{x} = 2x + k\), find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.
 
Re: Can't get this one started

Here is a big hint. Graph the expression using k <0, 0<=k<2, then 2<=k.
 
Re: Can't get this one started

masters said:
\(\displaystyle 4\cdot \sqrt{x} = 2x + k\), find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.
Click to expand...

Posting the same question in several places is NOT likely to win friends.

I just spent some time answering this same question you submitted to another service.

DON'T do that.

Oh...and it would be very nice if you showed some of YOUR work on your multiple postings....
 
Careful Mrspi, This is the first time I've posted this equation. Check the poster's name. This was part of a private message question I was trying to answer for another poster in another forum. Maybe they posted it. I thought I was on the right track, but wanted to confer with some of you guys. This is not a contest forum. Just trying to help folks.

I haven't seen your response. Don't even know where to look for it. But here's what I came up with.

\(\displaystyle 4\sqrt{x}=2x+k\)

Square both sides and set the equation in standard form: (I'll leave out the details)

\(\displaystyle 4x^2+(4k-16)x +k^2=0\)

Checking the discriminant (D), if D=0, then there is one real root, D>0, there are 2 real roots, and if D<0, there are no real roots.

I found D=-128k+256.

So, when D=0, k=2
When D>0, 0<k<2
When D<0, k>2
 
As I said above, work with the original expression.
Squaring both sides introduces possibly more roots.
Graph \(\displaystyle 2x - 4\sqrt x + k\) for the values of k that I suggested.
You will see an interesting pattern.
 
masters said:
Careful Mrspi, This is the first time I've posted this equation....I haven't seen your responce. Don't even know where to look for it.
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So... you don't know what Mrs. Pi is talking about, and have no idea where to look for her reply to another instance of this question-- but you're actively working on that other instance, conversing privately with the other poster(s)...?

I'm confused. :shock:

masters said:
This was part of a private message question I was trying to answer for another poster in another forum. Maybe they posted it. I thought I was on the right track, but wanted to confer with some of you guys....Just trying to help folks.
Click to expand...
If you're a tutor (or are currently acting as such), then you understand the importance of providing context ("This isn't my homework; I'm helping a student try to figure it out; he's studying X, has recently covered Y, and is familiar with Z...") and showing your work ("This is what he's covered recently in class; he tried this, and I've tried these; the book says the answer is as follows...").

The tutors can only respond to what is actually posted; they can't know what you don't provide. If you post the exact same text as was already answered elsewhere, the tutors are perfectly reasonable in assuming, based on years of experience, exactly what was assumed here.

Thank you for your understanding and for your future consideration.

Eliz.
 
Dear Masters,

To answer this last question, "Yes". Blatant truth-telling can be very useful. It's greatest potential is when it is received by the honest and sincere.

Let's learn some math, shall we?
 
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