#### hi2you

how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?

#### topsquark

Math Team
how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
$$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}}$$

Try multiplying the numerator and denominator by the conjugate:
$$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$

Can you finish it?

-Dan

1 person

#### hi2you

but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?

#### skeeter

Math Team
but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
much easier the way topsquark has recommended ... and that is working "from the back"

#### aurel5

but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?

$$\displaystyle \color{blue}{\sqrt3-\sqrt2=(\sqrt3-\sqrt2)\cdot \dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}= \dfrac{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{\sqrt3+\sqrt2}=\dfrac{( \sqrt3)^2-(\sqrt2)^2}{\sqrt3+\sqrt2} = \dfrac{3-2}{\sqrt3+\sqrt2} = \dfrac{1}{\sqrt3+\sqrt2}}$$

#### aurel5

how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?

Where is the logarithm ...?

#### hi2you

sorry! because this question was in my logarithm homework questions so I thought I needed to use logarithms! But thank you for your help !!

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