Question

Rationalize the denominator and simplify Root 3 minus root 2 / root 3 plus root 2

Answer 1

GIVEN :

Rationalize the denominator and simplify $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

TO SIMPLIFY :

The given expression $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

SOLUTION :

Given expression is $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Now rationalize the denominator of the given expression by multiplying and dividing the given expression by its conjugate we get,

$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})$

By using the algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$

By using the algebraic identity :

$(a-b)^2=a^2-2ab+b^2$

$=\frac{(\sqrt{3})^2-2(\sqrt{3})(\sqrt{2})+(\sqrt{2})^2}{3-2}$

By using the property:

$\sqrt{a}\times \sqrt{b}=\sqrt{ab}$

$=\frac{3-2\sqrt{6}+2}{1}$

$=5-2\sqrt{6}$

∴  $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=5-2\sqrt{6}$

Hence the given expression is rationalized and simplified as $5-2\sqrt{6}$        

Answer 2

Answer:

5 - 2 root 6 is the correct answer

Step-by-step explanation:

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