How do you rationalize $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

Question

How do you rationalize $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

1 Answer

Impact of this question

1820 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-08T16:37:53

7 $(\sqrt{3} + \sqrt{2})$ $(sqrt3+sqrt2)$

Multiply the numerator and the denominator by $\sqrt{3} + \sqrt{2}$ $sqrt3+sqrt2$ , that would make the denominator $(\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2})$ $(sqrt3-sqrt2)(sqrt3+sqrt2)$ =3-2= 1. The result would be $7 (\sqrt{3} + \sqrt{2})$ $7(sqrt3+sqrt2)$ . That is what was required.

Science

Math

Humanities

... and beyond

How do you rationalize $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you rationalize 7√3−√27/(sqrt3-sqrt2)?

1 Answer

Related questions

Impact of this question

How do you rationalize $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?