prasadini
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\sqrt{3} + \sqrt{2} / \sqrt{3} - \sqrt{2}
The discussion focuses on the rationalization of the expression $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$. Participants clarify that to rationalize the denominator, one must multiply by the conjugate, resulting in the expression $$\frac{(\sqrt{3}+\sqrt{2})^2}{3-2}$$, which simplifies to $$5 + 2\sqrt{6}$$. Additionally, the discussion outlines the method for rationalizing both the numerator and denominator, providing a general formula applicable to expressions of the form $$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$$.
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MarkFL said:Hello and welcome to MHB, prasadini! (Wave)
I've moved your thread to a more fitting area. :D
So, we are given the expressions (I assume):
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$
I assume you are to rationalize the denominator...what form of $1$ do we need to multiply this expression by to accomplish this?
prasadini said:$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$
is equal to
5+2√6 and 1 /5−2√6 How can i get this answer