Quick Question

  • Thread starter _Mayday_
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  • #1
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[SOLVED] Quick Question

Hey!

This question may tex ability :shy:

The Question
Write [tex]\frac{2\sqrt x+3}{x}[/tex] in the form [tex]2x^p + 3x^q[/tex] where p and q are constants.

Attempt :eek:

[tex]\frac{2\sqrt x+3}{x} \times x[/tex]

[tex]2x + 3x[/tex]

I basically multiplied the initial equation by x to get rid of the denominator, and now I have what looks like a close to correct answer, but without p and q, either p or q could be one but I think the fact that they have asked for p and q would suggest that they powers must be different.

Thanks to any helpers =]

_Mayday_
 

Answers and Replies

  • #2
Avodyne
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Can you write [itex]2\sqrt x/x[/itex] in the form [itex]2x^p[/itex]?

Can you write [itex]3/x[/itex] in the form [itex]3x^q[/itex]?

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).
 
  • #3
Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.
 
  • #4
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Can you write [itex]2\sqrt x/x[/itex] in the form [itex]2x^p[/itex]?

Can you write [itex]3/x[/itex] in the form [itex]3x^q[/itex]?

Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).

[tex]x^{\frac{1}{2}}/x[/tex]

[tex]3x^{-1}[/tex]

I don't think I got the first bit correct...

Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.

Thanks
 
  • #5
CompuChip
Science Advisor
Homework Helper
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Basic algebra:
[tex]
\frac{2\sqrt x+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2 \frac{\sqrt{x}}{x} + 3 \frac{1}{x}
[/tex]
 
  • #6
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I have the 3 but now, but I am still struggling with the root x. Please try and explain how it works.

Thank you very much for your help so far though, my maths is actually quite good but I have these little cracks I need filling in...

Thanks again.
 
  • #7
HallsofIvy
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You are correct that [itex]\sqrt{3}= 3^{1/2}[/itex]. Now what about the 1/x? What do the "laws of exponents" tell you about [itex]x^m/x^n[/itex]?
 
  • #8
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Okay, I will take your post above as an example HallsofIvy? I tried looking for the root 3 but couldn't find it in my working! :rofl:

I will take it that my answer is right:

[tex]x^\frac{1}{2}/x[/tex]

which gives me...

[tex]x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}[/tex]



Thanks Halls, I hope I am right...
 
  • #9
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Okay, I will take your post above as an example HallsofIvy? I tried looking for the root 3 but couldn't find it in my working! :rofl:

I will take it that my answer is right:

[tex]x^\frac{1}{2}/x[/tex]

which gives me...

[tex]x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}[/tex]



Thanks Halls, I hope I am right...

Eh, your answer is right but oddly derived.

[tex]\frac{x^\frac{1}{2}}{x}=x^{\frac{1}{2}-1}=x^{\frac{-1}{2}}[/tex]
 
  • #10
HallsofIvy
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It was just a typo. He meant
[tex]x^{\frac{1}{2}}\times x^{-1}= x^{-\frac{1}{2}}[/tex]
 
  • #11
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I appreciate all of your help everyone, thankyou!

_Mayday_
 

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