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Sladkaya [172]
1 year ago
9

4√b

Mathematics
1 answer:
Gemiola [76]1 year ago
7 0

Answer:

  \dfrac{4\sqrt{3b}+4b\sqrt{b}}{3-b^2}

Step-by-step explanation:

The factoring of the difference of squares is used to "rationalize the denominator" of rational functions having square roots in the denominator.

<h3>Difference of squares</h3>

The factoring of the difference of squares is ...

  a² -b² = (a -b)(a +b)

This means that an expression in which 'a' or 'b' or both are square roots can be made "rational" by making use of this relation.

<h3>Application</h3>

To eliminate the radical from the denominator, we multiply numerator and denominator by the "conjugate" of the denominator. The result is the denominator becomes a difference of squares, one of which is the square of the radical.

  \dfrac{4\sqrt{b}}{\sqrt{3}-b}=\dfrac{4\sqrt{b}}{\sqrt{3}-b}\cdot\dfrac{\sqrt{3}+b}{\sqrt{3}+b}=\boxed{\dfrac{4\sqrt{3b}+4b\sqrt{b}}{3-b^2}}

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