If \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} +… — JAMB UTME

If $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = k\sqrt{15}$ , find the value of $k$ .

Answer: 2

The value of

k

2

Combining the fractions gives a common denominator of

(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}) = 5-3 = 2

. Expanding the numerator gives

(8+2\sqrt{15}) - (8-2\sqrt{15}) = 4\sqrt{15}

. Dividing by 2 yields

2\sqrt{15}

, giving

k = 2

Combine the fractions on the left-hand side over a common denominator.

\frac{(\sqrt{5} + \sqrt{3})^2 - (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}

Combining two fractions with conjugate denominators simplifies the expression.

Evaluate the denominator using the difference of two squares formula

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

(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2

Multiplying conjugate surds eliminates the radical signs in the denominator.

Expand both squared terms in the numerator and subtract them.

(8 + 2\sqrt{15}) - (8 - 2\sqrt{15}) = 4\sqrt{15}

Expanding

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

gives

5 \pm 2\sqrt{15} + 3 = 8 \pm 2\sqrt{15}

Divide the resulting numerator by the denominator and solve for

k

\frac{4\sqrt{15}}{2} = 2\sqrt{15} \Rightarrow k = 2

Dividing

4\sqrt{15}

2

yields

2\sqrt{15}

, so matching the coefficients gives

k = 2

Rationalization of Denominators and Difference of Conjugate Surd Fractions

Estimated Time:1m 30s

Question