# Math Help - Quick conjugate clarifcation

1. ## Quick conjugate clarifcation

Consider the function $f(x)=(18x+18h)(\sqrt{15x-3})-18x\sqrt{15x+15h-3}$

If you factor out an $18x$, then theres still a $(1+18h)$ in front of the first square root. Would the conjugate just be the square roots or would it include the $(1+18h)$?

Thanks for the help!

Edit: Oops, I didn't mean to factor out the $18x$ but if there were still another term in front of the square root, would the same process apply?

2. Normally yes you can factorise. For example:

$2\sqrt{x+y} + 4\sqrt{x+z} = 2(\sqrt{x+y} + 2\sqrt{x+z})$

In your example you can still factor out 18

3. Originally Posted by e^(i*pi)
Normally yes you can factorise. For example:

$2\sqrt{x+y} + 4\sqrt{x+z} = 2(\sqrt{x+y} + 2\sqrt{x+z})$

In your example you can still factor out 18
Thanks! So it will only work assuming only one of the square roots has a term in front of it? If it were in the form of $a\sqrt{b}-c\sqrt{d}$, is there any conjugate for that? If not, are there any other ways to simplify it?

Edit: Wait, how is it possible to factor out an $18x$ from my original example?

Edit #2: Oh, you meant $18$. In that case, is there any way to simplify that radical? Because there are two different terms in front of them.

4. Is there more to this problem?

looks an awful lot like a miscalculated numerator of a difference quotient ...

$18(x+h)\sqrt{15x-3} - 18x\sqrt{15(x+h)-3}$

5. Originally Posted by youngb11
Thanks! So it will only work assuming only one of the square roots has a term in front of it? If it were in the form of $a\sqrt{b}-c\sqrt{d}$, is there any conjugate for that? If not, are there any other ways to simplify it?
There is a conjugate for every surd expression. It's just that this one will be a binomial: $a\sqrt{b} + c \sqrt{d}$. A conjugate can be obtained by changing the sign in the middle.

Edit #2: Oh, you meant $18$. In that case, is there any way to simplify that radical? Because there are two different terms in front of them.
Yes, if you look carefully you can also take out a $\sqrt{3}$ by using surd laws

$f(x) = 18(x+h\sqrt{3(5x-1)} - x\sqrt{3(5x+5h-1)}) = 18\sqrt{3}(x+h\sqrt{5x-1} - \sqrt{5x+5x-1})$

6. Originally Posted by skeeter
Is there more to this problem?

looks an awful lot like a miscalculated numerator of a difference quotient ...

$18(x+h)\sqrt{15x-3} - 18x\sqrt{15(x+h)-3}$
Actually, I have a test tomorrow involving differentiation using first principles (I hate those!) and I was trying to differentiate $f(x)=\frac{18x}{\sqrt{15x-3}}$

While doing the numerator, I got $\frac{18x+18h}{\sqrt{15x+15h-3}} - \frac{18x}{\sqrt{15x-3}}$

7. Leave it factored, it's much less trouble to work out the conjugate

$f'(x) = \displaystyle \lim_{h \to 0} \frac{18(x+h) - 18x}{\sqrt{15(x+h)-3} - \sqrt{15x-3}}$

Of course you may factor out $\sqrt{3}$ in the denominator

$\sqrt{15(x+h)-3} - \sqrt{15x-3} = \sqrt{3}\sqrt{5(x+h)-1} - \sqrt{3}\sqrt{5x-1}$

$= \sqrt{3}(\sqrt{5(x+h)-1} - \sqrt{5x-1})$

Hence the fraction is

$f'(x) = \displaystyle \lim_{h \to 0} \dfrac{18h}{\sqrt{3}(\sqrt{5(x+h)-1} - \sqrt{5x-1})}$

$= \displaystyle \lim_{h \to 0} \dfrac{6h\sqrt{3}}{\sqrt{5(x+h)-1} - \sqrt{5x-1}}$

The conjugate of the denominator is $\sqrt{5(x+h)-1} + \sqrt{5x-1}$

8. Originally Posted by e^(i*pi)
There is a conjugate for every surd expression. It's just that this one will be a binomial: $a\sqrt{b} + c \sqrt{d}$. A conjugate can be obtained by changing the sign in the middle.

Yes, if you look carefully you can also take out a $\sqrt{3}$ by using surd laws

$f(x) = 18(x+h\sqrt{3(5x-1)} - x\sqrt{3(5x+5h-1)}) = 18\sqrt{3}(x+h\sqrt{5x-1} - \sqrt{5x+5x-1})$
Aren't you missing an $x$ in front of $\sqrt{5x+5x-1}$?

And back to what you said, conjugate of $(18x+18h)(\sqrt{15x-3}-18x(\sqrt{15x+15h-3}$ would be $(18x+18h)(\sqrt{15x-3}+18x(\sqrt{15x+15h-3}$?

9. Erm, sorry for all the questions, but what do you mean leave it factored? And how did you get $\frac{18(x+h) - 18x}{\sqrt{15(x+h)-3} - \sqrt{15x-3}}$? Aren't they two different terms

Originally Posted by e^(i*pi)
Leave it factored, it's much less trouble to work out the conjugate

$f'(x) = \displaystyle \lim_{h \to 0} \frac{18(x+h) - 18x}{\sqrt{15(x+h)-3} - \sqrt{15x-3}}$

Of course you may factor out $\sqrt{3}$ in the denominator

$\sqrt{15(x+h)-3} - \sqrt{15x-3} = \sqrt{3}\sqrt{5(x+h)-1} - \sqrt{3}\sqrt{5x-1}$

$= \sqrt{3}(\sqrt{5(x+h)-1} - \sqrt{5x-1})$

Hence the fraction is

$f'(x) = \displaystyle \lim_{h \to 0} \dfrac{18h}{\sqrt{3}(\sqrt{5(x+h)-1} - \sqrt{5x-1})}$

$= \displaystyle \lim_{h \to 0} \dfrac{6h\sqrt{3}}{\sqrt{5(x+h)-1} - \sqrt{5x-1}}$

The conjugate of the denominator is $\sqrt{5(x+h)-1} + \sqrt{5x-1}$

10. I used the first principles of differentiation based upon the value of f(x) you posted in post 6.

11. Originally Posted by e^(i*pi)
I used the first principles of differentiation based upon the value of f(x) you posted in post 6.
Yes, but would it not be $\frac{18x+18h}{\sqrt{15x+15h-3}} - \frac{18x}{\sqrt{15x-3}}$ And then you'd need a common denominator?

12. No, the first principle of differentiation is $f'(x) \equiv \displaystyle \lim_{h \to 0} \left(\dfrac{f(x+h)-f(x)}{h}\right)$