Question

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Answer 1

Answer:

$\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})$

Step-by-step explanation:

Considering the radical expression

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

​$\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n$

$=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]$

Simplifying

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

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}$

$=\left(\left(3+3\sqrt{2}\right)^{\frac{1}{2}}\right)^2$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$=\left(3+3\sqrt{2}\right)^{\frac{1}{2}\cdot \:2}$

$=3+3\sqrt{2}$          As $\frac{1}{2}\cdot \:2=1$

So, putting $3+3\sqrt{2}$ into Equation [A]

$=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]$

As

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

So, Equation [A] becomes

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

$=4(3+3\sqrt{2})$

Therefore,

$\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})$

Keywords: radical expression

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