What is the simplified form of the expression \frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqr… — JAMB UTME

What is the simplified form of the expression $\frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ ?

The simplified form is

12 + 5\sqrt{6}

Multiplying both numerator and denominator by the conjugate of the denominator,

(\sqrt{3} + \sqrt{2})

, clears the square roots in the denominator (resulting in

3 - 2 = 1

). Expanding the numerator gives

3\sqrt{6} + 6 + 6 + 2\sqrt{6}

, which simplifies cleanly to

12 + 5\sqrt{6}

Identify the conjugate of the denominator

The conjugate of

(\sqrt{3} - \sqrt{2})

(\sqrt{3} + \sqrt{2})

To rationalize a binomial denominator of the form

(\sqrt{a} - \sqrt{b})

, multiply numerator and denominator by

(\sqrt{a} + \sqrt{b})

Multiply the numerator and denominator by the conjugate

\frac{(3\sqrt{2} + 2\sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}

This removes the radical terms from the denominator using the difference of squares identity.

Expand the numerator and simplify the denominator

Denominator:

(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

.
Numerator:

3\sqrt{2}\cdot\sqrt{3} + 3\sqrt{2}\cdot\sqrt{2} + 2\sqrt{3}\cdot\sqrt{3} + 2\sqrt{3}\cdot\sqrt{2} = 3\sqrt{6} + 6 + 6 + 2\sqrt{6}

Apply the distributive property and basic radical simplification rules.

Combine like terms in the numerator

(6 + 6) + (3\sqrt{6} + 2\sqrt{6}) = 12 + 5\sqrt{6}

Collect rational numbers together and like surd terms together.

Rationalization of Denominators with Binomial Surds

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