Question

Rationalise the denominator of



2+root 3
2- root 3​
​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{Rationalize\:\:denominator\:\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{7+4\sqrt{3}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\underline{\underline{\sf{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{2+\sqrt{3}}{2+\sqrt{3}}$

$=\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}$

$\bullet\: \sf{Solve\:\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}$

$\begin{aligned}&\text { Apply exponent rule: } a^{b} \cdot a^{c}=a^{b+c} \\\\&(2+\sqrt{3})(2+\sqrt{3})=(2+\sqrt{3})^{1+1}\\\\&=(2+\sqrt{3})^{1+1}\end{aligned}$

$=\left(2+\sqrt{3}\right)^2$

$=2^2+2\cdot \:2\sqrt{3}+\left(\sqrt{3}\right)^2$

$=7+4\sqrt{3}$

$\bullet\: \sf{Solve\:\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}$

$=2^2-\left(\sqrt{3}\right)^2$

$=4-3$

$=1$

$=\dfrac{7+4\sqrt{3}}{1}$

$\huge\boxed{\sf{=7+4\sqrt{3}}}$

Answer 2

Answer:

Please go through the attachment.