If \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} = p + q\sqrt{3}, where p and q are rati… — JAMB UTME

If $\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} = p + q\sqrt{3}$ , where $p$ and $q$ are rational numbers, what is the value of $p + q$ ?

The value of

p + q

is 3.

Multiplying both numerator and denominator by the conjugate

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

yields

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

. Matching terms with

p + q\sqrt{3}

gives

p = 2

and

q = 1

, so

p + q = 3

Multiply the numerator and denominator by the conjugate of the denominator,

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

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

To eliminate radicals from the denominator.

Expand both numerator and denominator.

Numerator:

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

. Denominator:

6 - 2 = 4

Using algebraic expansion

(a+b)^2 = a^2 + 2ab + b^2

and difference of two squares.

Simplify the resulting fraction.

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

Dividing each term in the numerator by 4.

Equate to

p + q\sqrt{3}

to determine

p

and

q

, then find

p + q

p = 2

q = 1 \implies p + q = 2 + 1 = 3

Comparing rational and irrational parts separately.

Rationalization of Denominators and Surd Conjugates

Question