If \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} +… — JAMB UTME

If $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = k\sqrt{6}$ , find the value of the integer $k$ .

Answer: 4

Answer

The value of the integer

k

is 4.

Combining the fractions over the common denominator

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

yields a numerator of

(3+2+2\sqrt{6}) - (3+2-2\sqrt{6}) = 4\sqrt{6}

. Thus,

k\sqrt{6} = 4\sqrt{6}

, which gives

k = 4

Step-by-Step Solution

Combine the fractions using their common denominator

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

Subtracting algebraic fractions requires finding the least common denominator, which is the product of the conjugate pair.

Expand the terms in the numerator

(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \text{ and } (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}

Use the perfect square expansion formula

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

Subtract the expanded terms in the numerator and simplify the denominator

\text{Numerator: } (5 + 2\sqrt{6}) - (5 - 2\sqrt{6}) = 4\sqrt{6}, \text{ Denominator: } (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

Apply the difference of two squares identity

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

to the denominator and carefully distribute the negative sign across terms in the numerator.

Equate the simplified expression to

k\sqrt{6}

and solve for

k

k = 4

Comparing

\frac{4\sqrt{6}}{1} = 4\sqrt{6}

with

k\sqrt{6}

yields

k = 4

Key Concept

Rationalisation of surd denominators using conjugate pairs and difference of squares

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