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Mathematics 6 Online
OpenStudy (anonymous):

x=sqrt(7+4sqrt3)+sqrt(7-4sqrt3)

OpenStudy (anonymous):

OpenStudy (anonymous):

You may see the equation here.

OpenStudy (ash2326):

We have \[x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3}\] Let's multiply and divide this by \(\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}\) we have \[x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3} \times \frac{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\] We get \[x=\frac{{(\sqrt{7+4\sqrt 3})^2 }-(\sqrt{(7-4\sqrt 3})^2 }{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\] we get \[x=\frac{7+4\sqrt 3-(7-4 \sqrt 3)}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\] so we get \[x=\frac{8\sqrt 3}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]

OpenStudy (anonymous):

Is this the answer?

OpenStudy (ash2326):

Yeah can't be simplified more

OpenStudy (anonymous):

How it be made?

OpenStudy (anonymous):

Because the answer is 4.

OpenStudy (anonymous):

Ai Se Eu Te Pego -

OpenStudy (anonymous):

159

OpenStudy (anonymous):

rrsrsrs

OpenStudy (anonymous):

good

OpenStudy (anonymous):

De onde vc é Elodi?

OpenStudy (anonymous):

quoi

OpenStudy (anonymous):

Vc falou português

OpenStudy (anonymous):

i m from Brazil

OpenStudy (anonymous):

okk

OpenStudy (anonymous):

or u may send me your link profile

OpenStudy (ash2326):

I got it, Let me show you how this 4

OpenStudy (anonymous):

yes]

OpenStudy (ash2326):

@viniterranova please don't share personal info like facebook id here. Delete it please

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Bye Elodi See u around i ve got go

OpenStudy (anonymous):

y

OpenStudy (ash2326):

We have \[x=\sqrt {7+4 \sqrt 3}+\sqrt {7-4 \sqrt 3}\] Let's square both the sides we get \[x^2=(\sqrt {7+4 \sqrt 3})^2+(\sqrt {7-4 \sqrt 3})^2+2\times (\sqrt {7+4 \sqrt 3})\times (\sqrt {7-4 \sqrt 3})\] we get \[x^2=7+4\sqrt 3+7-4\sqrt 3+2 \sqrt{(7^2-(4\sqrt 3)^2}\] we get \[x^2=14+2\times \sqrt {49-16\times 3}\] we get \[x^2=14+2\times \sqrt 1\] we get \[x^2=14+2=16\] so \[x=4\]

OpenStudy (anonymous):

square on both sides|dw:1330781973743:dw|

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