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What is the conjugate? $\sqrt x +2 \sqrt b$ I'm not sure how to find the conjugate of any term. Help, please?

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You have a number $\sqrt{x}+2\sqrt{b}$ and you're asking for the conjugate.

Most times this refers to the complex conjugate, which involves replacing the imaginary part of the complex number with its negative.

Your number is a purely real number, so its complex conjugate is itself.

However, you may be wanting to rationalize a denominator of a fraction, in which case perhaps what you're looking for is the same expression with the second term having opposite sign, also called the binomial conjugate, so you can multiply top and bottom by it to get rid of the square roots in the bottom?

$$\frac{1}{\sqrt{x} + 2 \sqrt{b}} \times \frac{\sqrt{x} - 2 \sqrt{b}}{\sqrt{x} - 2 \sqrt{b}}= \frac{\sqrt{x} + 2 \sqrt{b}}{x + 4b}.$$

In this case, the conjugate involves switching the sign of the second term: $\sqrt{x}-2\sqrt{b}$.

(Note that the value of the conjugate depends on how it's written. If it were written $2\sqrt{b}+\sqrt{x}$, the conjugate of this expression would be $2\sqrt{b}-\sqrt{x}$.)

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  • $\begingroup$ What if i just needed the conjugate of the equation? Somebody said that you just switch the signs... Is that true? $\endgroup$
    – Tiffany
    Commented Jan 21, 2014 at 15:15
  • $\begingroup$ That's likely what you need, yes. That's what I did in the example above. You can rationalize the denominator using the difference of squares by multiplying the denominator by its conjugate. $\endgroup$
    – John
    Commented Jan 21, 2014 at 15:46
  • $\begingroup$ Added information to my answer. $\endgroup$
    – John
    Commented Jan 21, 2014 at 15:53
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Using the conjugate we switch the sign in between the two terms $\sqrt{x}+2\sqrt{b}$. We do this to create a difference of squares. The difference of squares can be seen in this example: $$(a+b)(a-b)=a^2-b^2$$ Notice how we don't have a middle term. This is intentional and the result of using the difference of squares.

$$=\left(\dfrac{1}{\sqrt{x} + 2 \sqrt{b}}\right)\cdot\left(\dfrac{\sqrt{x}-2\sqrt{b}}{\sqrt{x} - 2 \sqrt{b}}\right) $$

$$=\dfrac{1\left(\sqrt{x}-2\sqrt{b}\right)}{\left(\sqrt{x}+2\sqrt{b}\right)\left(\sqrt{x}-2\sqrt{b}\right)}$$

$$=\boxed{\dfrac{\sqrt{x}-2\sqrt{b}}{x-4b}}$$

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