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Rationalise the following:
$\frac{3+√5}{3-√5}$
Given: $\frac{3+√5}{3-√5}$
To do: Rationalize the given number
Solution:
$\frac{3+√5}{3-√5}$
[Multiply and divide by $3+\sqrt{5}$]
=$\frac{3+√5}{3-√5} \times \frac{3+√5}{3+√5}$
Numerator- $(a+b)(a+b)=(a+b)^2$
Denominator- $(a-b)(a+b)=a^2-b^2$
=$\frac{(3+√5)^{2}}{3^{2}-√5^{2}}$
= $\frac{3^{2} + 2 \times 3 \times √5 + (√5)^2} {9 - 5}$
= $\frac{9 +6√5 + 5}{4}$
= $\frac{14 + 6√5}{4} $
= $\frac{7+3√5}{2}$ Ans
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