# Quick conjugate clarifcation

• Feb 23rd 2011, 01:33 PM
youngb11
Quick conjugate clarifcation
Consider the function $\displaystyle f(x)=(18x+18h)(\sqrt{15x-3})-18x\sqrt{15x+15h-3}$

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

Thanks for the help!

Edit: Oops, I didn't mean to factor out the $\displaystyle 18x$ but if there were still another term in front of the square root, would the same process apply?
• Feb 23rd 2011, 01:40 PM
e^(i*pi)
Normally yes you can factorise. For example:

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

In your example you can still factor out 18
• Feb 23rd 2011, 01:43 PM
youngb11
Quote:

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

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

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 $\displaystyle a\sqrt{b}-c\sqrt{d}$, 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 $\displaystyle 18x$ from my original example?

Edit #2: Oh, you meant $\displaystyle 18$. In that case, is there any way to simplify that radical? Because there are two different terms in front of them.
• Feb 23rd 2011, 01:51 PM
skeeter
Is there more to this problem?

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

$\displaystyle 18(x+h)\sqrt{15x-3} - 18x\sqrt{15(x+h)-3}$
• Feb 23rd 2011, 01:52 PM
e^(i*pi)
Quote:

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 $\displaystyle a\sqrt{b}-c\sqrt{d}$, 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: $\displaystyle a\sqrt{b} + c \sqrt{d}$. A conjugate can be obtained by changing the sign in the middle.

Quote:

Edit #2: Oh, you meant $\displaystyle 18$. 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 $\displaystyle \sqrt{3}$ by using surd laws

$\displaystyle f(x) = 18(x+h\sqrt{3(5x-1)} - x\sqrt{3(5x+5h-1)}) = 18\sqrt{3}(x+h\sqrt{5x-1} - \sqrt{5x+5x-1})$
• Feb 23rd 2011, 01:55 PM
youngb11
Quote:

Originally Posted by skeeter
Is there more to this problem?

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

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

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

While doing the numerator, I got $\displaystyle \frac{18x+18h}{\sqrt{15x+15h-3}} - \frac{18x}{\sqrt{15x-3}}$
• Feb 23rd 2011, 02:08 PM
e^(i*pi)
Leave it factored, it's much less trouble to work out the conjugate

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

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

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

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

Hence the fraction is

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

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

The conjugate of the denominator is $\displaystyle \sqrt{5(x+h)-1} + \sqrt{5x-1}$
• Feb 23rd 2011, 02:11 PM
youngb11
Quote:

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

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

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

Aren't you missing an $\displaystyle x$ in front of $\displaystyle \sqrt{5x+5x-1}$?

And back to what you said, conjugate of $\displaystyle (18x+18h)(\sqrt{15x-3}-18x(\sqrt{15x+15h-3}$ would be $\displaystyle (18x+18h)(\sqrt{15x-3}+18x(\sqrt{15x+15h-3}$?
• Feb 23rd 2011, 02:18 PM
youngb11
Erm, sorry for all the questions, but what do you mean leave it factored? And how did you get $\displaystyle \frac{18(x+h) - 18x}{\sqrt{15(x+h)-3} - \sqrt{15x-3}}$? Aren't they two different terms(Worried)

Quote:

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

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

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

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

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

Hence the fraction is

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

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

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

• Feb 23rd 2011, 02:24 PM
e^(i*pi)
I used the first principles of differentiation based upon the value of f(x) you posted in post 6.
• Feb 23rd 2011, 02:27 PM
youngb11
Quote:

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 $\displaystyle \frac{18x+18h}{\sqrt{15x+15h-3}} - \frac{18x}{\sqrt{15x-3}}$ And then you'd need a common denominator?
• Feb 23rd 2011, 02:37 PM
e^(i*pi)
No, the first principle of differentiation is $\displaystyle f'(x) \equiv \displaystyle \lim_{h \to 0} \left(\dfrac{f(x+h)-f(x)}{h}\right)$