What is the simplified form of the expression \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{… — JAMB UTME

What is the simplified form of the expression $\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} - \sqrt{3}$ ?

The simplified form of the expression is

2

Multiplying the numerator and denominator of

\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}

by the conjugate

(\sqrt{6} + \sqrt{2})

gives

\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}

. Subtracting

\sqrt{3}

yields

2

Rationalize the denominator of

\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}

Multiply numerator and denominator by the conjugate

(\sqrt{6} + \sqrt{2})

to get

\frac{(\sqrt{6} + \sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}

Rationalizing eliminates the surd from the denominator using the difference of two squares identity.

Expand the numerator and simplify the fraction

Numerator:

(\sqrt{6})^2 + 2\sqrt{12} + (\sqrt{2})^2 = 6 + 4\sqrt{3} + 2 = 8 + 4\sqrt{3}

. Denominator:

6 - 2 = 4

. Fraction simplifies to

\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}

Simplifying radical factors (

\sqrt{12} = 2\sqrt{3}

) allows division by the common denominator.

Subtract

\sqrt{3}

from the rationalized term

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

Subtracting like surd terms cancels out

\sqrt{3}

leaving the rational constant.

Rationalization of Binomial Denominators and Surd Simplification

Soru