Question

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Rationalise the denominator of
$\frac{a + \sqrt{4b} }{a - \sqrt{4b} }$
where a is an integer and b is a prime number.
thanks. ​​

Answer 1

Answer:

$\dfrac{(a+2\sqrt{b})^2 }{a^2-4b}$

Step-by-step explanation:

To rationalise the denominator, multiply by the expression by $\dfrac{a+\sqrt{4b} }{a+\sqrt{4b} }$:

$=\dfrac{a+\sqrt{4b} }{a-\sqrt{4b} }\times\dfrac{a+\sqrt{4b} }{a+\sqrt{4b} }$

$=\dfrac{(a+\sqrt{4b}) (a+\sqrt{4b})}{(a-\sqrt{4b}) (a+\sqrt{4b})}$

$=\dfrac{(a+\sqrt{4b})^2 }{a^2-4b}$

$=\dfrac{(a+2\sqrt{b})^2 }{a^2-4b}$

Answer 2

$\\ \rm\hookrightarrow \dfrac{a+\sqrt{4b}}{a-\sqrt{4b}}$

• √4b be x

$\\ \rm\hookrightarrow \dfrac{a+x}{a-x}$

$\\ \rm\hookrightarrow \dfrac{(a+x)^2}{a^2-x^2}$

$\\ \rm\hookrightarrow \dfrac{a^2+2ax+x^2}{a^2-x^2}$

• x^2=√4b^2=4b

$\\ \rm\hookrightarrow \dfrac{a^2+2a\sqrt{4b}+4b}{a^2-4b}$