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I'm having problems finding the conjugate of $\frac{1}{4\sqrt{3}-7}$

The answer I get is as follows: ${-4\sqrt{3}-7}$

However the answer given is ${-4\sqrt{3}+7}$

Here are my workings out... $\frac{1}{4\sqrt{3}-7} * \frac{4\sqrt{3}+7}{4\sqrt{3}+7} = \frac{4\sqrt{3}+7}{-1} = {-4\sqrt{3}-7}$

I'd appreciate any guidance on where I've gone wrong.

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    $\begingroup$ How did you get $-1$ for the denominator? And what is your definition of "conjugate"? $\endgroup$
    – Eric Wofsey
    Commented Feb 12, 2017 at 23:17
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    $\begingroup$ In beginning you say the answer you got is $-4(\sqrt{3}-7)$ but when you show your working you get $-4(\sqrt{3}+7)$ $\endgroup$
    – kingW3
    Commented Feb 12, 2017 at 23:20
  • $\begingroup$ Aren't you forgetting a square root for that $\;7\;$ ? $\endgroup$
    – DonAntonio
    Commented Feb 12, 2017 at 23:21
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    $\begingroup$ Wait:$$(\sqrt3-7)(\sqrt3+7)=3-49=-46\implies \frac1{4(\sqrt3)}=\frac{\sqrt3+7}{-46}\ldots$$ How is there nothing wrong with his work?? Am I missing something here? $\endgroup$
    – DonAntonio
    Commented Feb 12, 2017 at 23:23
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    $\begingroup$ @Hemmed Yes...brackets/parentheses in mathematics can make a huge difference...! $\endgroup$
    – DonAntonio
    Commented Feb 12, 2017 at 23:27

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Well, after all the changes:

$$(4\sqrt3-7)(4\sqrt3+7)=48-49=-1$$

and then indeed:

$$\frac1{4\sqrt3-7}=-\left(4\sqrt3+7\right)=-4\sqrt3-7$$

and yes: the conjugate is $\;\frac1{4\sqrt3+7}\;$ , and what you say is the answer given is neither the conjugate nor the product of the original expression by its conjugate.

Added following an idea by projectilemotion: Perhaps the idea is first to rationalize the expression and then to find its conjugate:

$$\text{Rationalizing:}\;\;\frac1{4\sqrt3-7}\cdot\frac{4\sqrt3+7}{4\sqrt3+7}=-4\sqrt3-7$$

and now the rightmost expression's conjugate indeed is: $\;-4\sqrt3\color{Red}+7\;$ .....tadaaah!

It's hard to tell what they meant without knowing a priori their definitions... BTW, in this case and for me, the conjugate could as well be $\;4\sqrt3-7=-7+4\sqrt3\;$ ...Funny.

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  • $\begingroup$ What I don't understand is that the conjugate I get ends with -7 but in the example given in the quiz the conjugate is actually +7. $\endgroup$
    – Hemmed
    Commented Feb 12, 2017 at 23:29
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    $\begingroup$ @projectilemotion That seems to be a reasonable assumption. I couldn't tell without knowing the proper definition the use. $\endgroup$
    – DonAntonio
    Commented Feb 12, 2017 at 23:40
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    $\begingroup$ @Hemmed Read the stuff I just added to my answer. It fits what you say the answer and it is most probably they way they defined in your school the term "conjugate": after first rationalizing and etc. $\endgroup$
    – DonAntonio
    Commented Feb 12, 2017 at 23:48
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    $\begingroup$ @Hemmed ?? Well, that is basic mathematics: $$\;-a-b=-(a+b)=\frac{-(a+b)}1=\frac{a+b}{-1}=\ldots etc.$$ $\endgroup$
    – DonAntonio
    Commented Feb 13, 2017 at 0:12
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    $\begingroup$ @DonAntonio I had a brain fail :( $\endgroup$
    – Hemmed
    Commented Feb 13, 2017 at 0:14

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