# Quick Question

[SOLVED] Quick Question

Hey!

This question may tex ability :shy:

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

Attempt

$$\frac{2\sqrt x+3}{x} \times x$$

$$2x + 3x$$

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

Avodyne
Can you write $2\sqrt x/x$ in the form $2x^p$?

Can you write $3/x$ in the form $3x^q$?

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

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.

Can you write $2\sqrt x/x$ in the form $2x^p$?

Can you write $3/x$ in the form $3x^q$?

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

$$x^{\frac{1}{2}}/x$$

$$3x^{-1}$$

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

CompuChip
Homework Helper
Basic algebra:
$$\frac{2\sqrt x+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2 \frac{\sqrt{x}}{x} + 3 \frac{1}{x}$$

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.

HallsofIvy
Homework Helper
You are correct that $\sqrt{3}= 3^{1/2}$. Now what about the 1/x? What do the "laws of exponents" tell you about $x^m/x^n$?

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:

$$x^\frac{1}{2}/x$$

which gives me...

$$x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}$$

Thanks Halls, I hope I am right...

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:

$$x^\frac{1}{2}/x$$

which gives me...

$$x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}$$

Thanks Halls, I hope I am right...

Eh, your answer is right but oddly derived.

$$\frac{x^\frac{1}{2}}{x}=x^{\frac{1}{2}-1}=x^{\frac{-1}{2}}$$

HallsofIvy
$$x^{\frac{1}{2}}\times x^{-1}= x^{-\frac{1}{2}}$$