anonymous
  • anonymous
x=sqrt(7+4sqrt3)+sqrt(7-4sqrt3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
You may see the equation here.
ash2326
  • 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 \[x=\frac{8\sqrt 3}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]

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anonymous
  • anonymous
Is this the answer?
ash2326
  • ash2326
Yeah can't be simplified more
anonymous
  • anonymous
How it be made?
anonymous
  • anonymous
Because the answer is 4.
anonymous
  • anonymous
Ai Se Eu Te Pego -
anonymous
  • anonymous
159
anonymous
  • anonymous
rrsrsrs
anonymous
  • anonymous
good
anonymous
  • anonymous
De onde vc é Elodi?
anonymous
  • anonymous
quoi
anonymous
  • anonymous
Vc falou português
anonymous
  • anonymous
i m from Brazil
anonymous
  • anonymous
okk
anonymous
  • anonymous
or u may send me your link profile
ash2326
  • ash2326
I got it, Let me show you how this 4
anonymous
  • anonymous
yes]
ash2326
  • ash2326
@viniterranova please don't share personal info like facebook id here. Delete it please
anonymous
  • anonymous
ok
anonymous
  • anonymous
Bye Elodi See u around i ve got go
anonymous
  • anonymous
y
ash2326
  • 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 \[x=4\]
anonymous
  • anonymous
square on both sides|dw:1330781973743:dw|
anonymous
  • anonymous
Good.

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