Given that \frac{\sqrt{5} + 2}{\sqrt{5} - 2} = a + b\sqrt{5}, where a and b are rational n… — JAMB UTME

Given that $\frac{\sqrt{5} + 2}{\sqrt{5} - 2} = a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers, what is the value of $a + b$ ?

The value of

a + b

is 13.

To simplify

\frac{\sqrt{5} + 2}{\sqrt{5} - 2}

, multiply both numerator and denominator by the conjugate of the denominator, which is

\sqrt{5} + 2

. Expanding the numerator

(\sqrt{5} + 2)^2

gives

5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}

. The denominator simplifies to

(\sqrt{5})^2 - 2^2 = 5 - 4 = 1

. Thus, the expression becomes

9 + 4\sqrt{5}

. Comparing this to

a + b\sqrt{5}

gives

a = 9

and

b = 4

. Therefore,

a + b = 13

Identify the conjugate of the denominator

The denominator is

\sqrt{5} - 2

, so its conjugate is

\sqrt{5} + 2

Multiplying by the conjugate rationalizes the binomial denominator using the difference of two squares.

Multiply the numerator and denominator by the conjugate

\frac{(\sqrt{5} + 2)(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{(\sqrt{5} + 2)^2}{(\sqrt{5})^2 - 2^2}

This removes the radical from the denominator.

Expand both numerator and denominator

\frac{5 + 4\sqrt{5} + 4}{5 - 4} = \frac{9 + 4\sqrt{5}}{1} = 9 + 4\sqrt{5}

Using

(x + y)^2 = x^2 + 2xy + y^2

for the numerator and

(x - y)(x + y) = x^2 - y^2

for the denominator.

Equate to

a + b\sqrt{5}

and solve for

a + b

a = 9

and

b = 4

, so

a + b = 9 + 4 = 13

Matching the rational part

a

and coefficient of the surd

b

gives the target sum.

Rationalization of Binomial Denominators

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